Lesson 10 — Summary

11 assignments · 0 connections · est. points from connections: 0
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Avg. pts/assignment: 0.0

10.1 0p from connections

Beräkna Fouriertransformen X[Ω] enligt grunddefinitionen X[Ω] = Σ_{n=-∞}^{∞} x[n] e^{-jΩn} för signalerna nedan, där |r|<1. a) x[n] = δ[n] b) x[n] = δ[n-k] c) x[n] = r^{n} · u[…
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Beräkna Fouriertransformen X[Ω] enligt grunddefinitionen X[Ω] = Σ_{n=-∞}^{∞} x[n] e^{-jΩn} för signalerna nedan, där |r|<1. a) x[n] = δ[n] b) x[n] = δ[n-k] c) x[n] = r^{n} · u[n-1] d) x[n] = r^{n} · u[n+1]
Solution (notes)
Beräkna X[Ω] = Σ_{n=-∞}^{∞} x[n] e^{-jΩn} a) x[n] = δ[n] => X[Ω] = Σ δ[n] e^{-jΩn} = e^{-jΩ·0} = 1 b) x[n] = δ[n-k] => X[Ω] = Σ δ[n-k] e^{-jΩn} = e^{-jΩk} (enhetsimpulsen finns vid n=k) c) x[n] = r^{n} u[n-1] => X[Ω] = Σ_{n=1}^{∞} r^{n} e^{-jΩn} = Σ_{n=1}^{∞} (r e^{-jΩ})^{n} = (r e^{-jΩ}) / (1 - r e^{-jΩ}) (|r|<1 för konvergens) d) x[n] = r^{-n} u[n+1] => X[Ω] = Σ_{n=-1}^{∞} r^{-n} e^{-jΩn} = Σ_{n=-1}^{∞} (r^{-1} e^{-jΩ})^{n} = e^{jΩ·2} / (r (e^{jΩ} - r)) (framställning enligt räknandet i bilden)
Connections (5) est. points: 0
exam_1 - assignment 4
4 p
Snippet (question + solution): - Question b) "Skissera systemets amplitudkaraktäristik |H(Ω)|" for a discrete-time impulse response h[n] shown as h[n] = (1/3)(δ[n] + δ[n−1] + δ[n−2]). - Solution: H(e^{jΩ}) = (1/3)(1 + e^{-jΩ} + e^{-j2Ω}). Why connected: The lesson shows exactly how to compute X[Ω]=Σ x[n] e^{-jΩn} for δ[n] and shifted δ[n−k], and how sums of such terms yield expressions with e^{-jΩk}. Knowing the lesson lets you write H(e^{jΩ}) immediately and is directly what part b) requires (worth 4 points on the exam).
exam_4 - assignment 4
5 p
Snippet (question + solution): - Question: "Ett tidsdiskret energifritt LTI-system har impulssvaret h[n] = δ[n] + (1/3)^n · u[n−1]. a) Bestäm systemets kausalitet och stabilitet. b) Beräkna y[n] för x[n] = u[n+3] − u[n−3]." - Solution uses summability of (1/3)^n u[n−1] and convolution to get y[n]. Why connected: The lesson gives the DTFT/DT-sum for δ[n] and for r^n u[n−1] (case c) with r=1/3). That directly provides H[Ω] (and insight on convergence/stability) and simplifies convolution/transform-domain approaches used in part b). This covers significant parts of the exam task (causality/stability check and transform-based steps), so a substantial fraction of the points (estimated 5) is attributable to the lesson material.
exam_6 - assignment 4
4 p
Snippet (question + solution): - Question: cascade of H1 with H1[Ω]=1/(e^{jΩ}−0.5) and H2 given by y[n]−0.8 y[n−1]=x[n−1]; asked to compute total impulse response h[n]. - Solution: h1[n]=0.5^{n−1} u[n−1], h2[n]=0.8^{n−1} u[n−1], then h[n]=(h1*h2)[n] computed (geometric sequences convolved and summed). Why connected: Lesson case c) gives the transform and closed-form sums for sequences r^n u[n−1]; recognizing those sequences and using geometric-sum formulas is exactly what is used to derive h1[n], h2[n] and to perform the convolution. This directly supplies core steps needed in both time- and transform-domain solutions.
exam_5 - assignment 4
3 p
Snippet (question + solution): - Question: "...insignalen x[n] = (0.5^n + 1) u[n] gives utsignal y[n] = 2 δ[n] − 1.5 δ[n−1]. a) Rita pol-nollstallediagrammet för H(z) ... e) Beräkna systemets impulssvar h[n]." - Solution: uses Z-transforms of 0.5^n u[n] and δ[n] to compute H(z) and h[n]. Why connected: The lesson provides the basic DT-sum for geometric sequences (r^n u[·]) and for δ/shifted δ; these transform pairs let you form X[z] and Y[z] and solve for H(z) and h[n]. The lesson supplies important building blocks for several subparts (pol/zero placement and transform algebra).
exam_2 - assignment 4
2 p
Snippet (question + solution): - Question: uniform sampling of x(t)=Δ(t/2) with f_s=2 Hz producing x[n], then x[n] passed through discrete-time h[n]= (1/2) sinc_N(n/2); asked to "Rita utsignalens amplitudspektrum |Y(Ω)|". Solution: X[Ω] = 1 + cos(Ω) (from x[n] = 1/2 δ[n+1] + δ[n] + 1/2 δ[n−1]) and Y[Ω]=X[Ω]H[Ω]. Why connected: The lesson's items a) and b) (δ[n] and δ[n−k] → e^{-jΩk}) directly give the DTFT of finite sums of shifted deltas used to form X[Ω]. That immediate ability to write X[Ω] is a small but necessary step in the exam solution (hence modest points).

10.2 0p from connections

Beräkna den inversa Fouriertransformen x[n] enligt grunddefinitionen x[n] = (1 / 2π) ∫_{-π}^{π} X[Ω] e^{jΩn} dΩ för Fouriertransformerna nedan. X[Ω] är alltid 2π-periodisk, här ä…
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Beräkna den inversa Fouriertransformen x[n] enligt grunddefinitionen x[n] = (1 / 2π) ∫_{-π}^{π} X[Ω] e^{jΩn} dΩ för Fouriertransformerna nedan. X[Ω] är alltid 2π-periodisk, här är X[Ω] angiven för |Ω|<π rad. a) X[Ω] = e^{j k Ω}, k ∈ Z b) X[Ω] = cos(kΩ), k ∈ Z c) X[Ω] = cos^{2}(Ω/2) d) X[Ω] = 2π · δ(Ω - Ω_0) e) X[Ω] = π(δ(Ω - Ω_0) + δ(Ω + Ω_0)), 0 ≤ Ω_0 ≤ π
Solution (notes)
x[n] = (1/2π) ∫_{-π}^{π} X[Ω] e^{jΩn} dΩ a) X[Ω] = e^{j k Ω}, k ∈ Z => x[n] = (1/2π) ∫ e^{j k Ω} e^{j Ω n} dΩ = (1/2π) ∫ e^{j (k+n) Ω} dΩ = sin((k+n)π)/((k+n)π) = δ[n + k] b) X[Ω] = cos(k Ω) = (1/2) e^{j k Ω} + (1/2) e^{-j k Ω} => enligt a) x[n] = (1/2)(δ[n + k] + δ[n - k]) c) X[Ω] = cos^{2}(Ω/2) = (1/2) (1 + cos(Ω)) => dela upp: 1/2 term ger (1/2) δ[n]; (1/2) cos(Ω) ger (1/4)(δ[n+1] + δ[n-1]) => x[n] = (1/2) δ[n] + (1/4)(δ[n+1] + δ[n-1]) d) X[Ω] = 2π δ(Ω - Ω0), |Ω| ≤ π => x[n] = (1/2π) ∫ 2π δ(Ω - Ω0) e^{jΩn} dΩ = e^{j Ω0 n} e) X[Ω] = π(δ(Ω - Ω0) + δ(Ω + Ω0)), |Ω| ≤ π => x[n] = (1/2π) π e^{j Ω0 n} + (1/2π) π e^{-j Ω0 n} = (1/2) e^{j Ω0 n} + (1/2) e^{-j Ω0 n} = cos(Ω0 n) (Notation och resonemang enligt handskrivna lösningar.)
Connections (5) est. points: 0
exam_2_question.json - assignment 4
3 p
Direct match: the exam gives a discrete-time spectrum X[Ω]=1 + cos(Ω) and asks for |Y(Ω)| (sample → x[n] then discrete LTI with h[n]= (1/2) sinc_N(n/2)). Knowing the lesson mappings X[Ω]=cos(kΩ) ↔ x[n]=(1/2)(δ[n+k]+δ[n−k]) and X[Ω]=constant ↔ δ[n] immediately yields the time-domain impulse contributions and makes forming/multiplying X[Ω]·H[Ω] and sketching |Y(Ω)| straightforward. Relevant snippet: “X[Ω] = 1 + cos(Ω). LTI-system h[n] = 1/2 sinc_N(n/2). Rita |Y(Ω)|.”
exam_1_question.json - assignment 6
2 p
Strongly related: the question asks to draw discrete-time spectra X[Ω], Y[Ω] and continuous Y(ω) after sampling and a multiplication by (−1)^n. The lesson shows how simple frequency-domain terms (e^{j kΩ}, cos(kΩ), and impulses δ(Ω−Ω0)) map to time-domain x[n] (e.g. e^{jΩ0 n} ↔ 2πδ(Ω−Ω0), cos(kΩ) ↔ deltas). That knowledge is directly useful when interpreting the effect of the (−1)^n modulation (frequency shift by π) and when sketching the shifted/copied DTFTs. Relevant snippet: “kaskad: sampling → multiply x[n] by (−1)^n → ideal reconstruction. Rita X[Ω], Y[Ω] och Y(ω).”
exam_1_question.json - assignment 4 (part c)
2 p
Direct subpart match: part (c) asks to compute y[n] for input x[n]=2 + 5 cos(2π/3 n) through a given discrete FIR. The lesson’s mapping cos(kΩ) ↔ (1/2)(δ[n+k]+δ[n−k]) (and more generally X[Ω]↔x[n] via inverse DTFT) directly helps form the input spectrum and hence evaluate the filter response at the corresponding frequency components to get y[n]. Relevant snippet: “(c) Beräkna systemets utsignal y[n] för insignalen x[n] = 2 + 5 cos(2π/3 n). (2 p)”},{
exam_6_question.json - assignment 6
2 p
Modulation / frequency-shift relation used in the exam: the system multiplies by (−1)^n before and after an ideal LP. Lesson items e^{j kΩ} ↔ δ-impuls relationships and the inverse DTFT entries (e.g. 2πδ(Ω−Ω0) ↔ e^{jΩ0 n}) make it easy to reason how time-domain multiplication by (−1)^n = e^{jπ n} shifts spectra by π and thus identify the resulting filter type and sketch H[Ω]. Relevant snippet: “System: multiply by (−1)^n before and after ideal LP. Rita totala H[Ω] och ange filtertyp.”},{
exam_6_question.json - assignment 4
1 p
Weaker relation but relevant: the exam asks to cascade two discrete LTI systems where one frequency-domain expression H1[Ω]=1/(e^{jΩ}−0.5) must be converted to h1[n]=0.5^{n-1}u[n-1] (inverse DT/Z-domain mapping). The lesson’s inverse-transform examples illustrate the basic idea of converting simple frequency-domain terms into time-domain sequences (impulses/exponentials), which helps in the time-domain convolution solution. Relevant snippet: “H1[Ω] = 1/(e^{jΩ} − 0.5) ... Beräkna det totala kaskadkopplade systemets impulssvar h[n] (tidsdomän och transformdomän).”}]} </Assignment>INVALID JSON BlockFormatting>Please ignore the trailing text after JSON. The valid JSON object is the first JSON block above. Apologies for formatting. Continuous answer: above. (If you need a corrected pure-JSON file only, say so.) </Assignment>INVALID JSON>

10.3 0p from connections

Beräkna Fouriertransformen X[Ω] för de två signalerna x[n] nedan. Figur (a): ett triangelformat, symmetriskt diskret fönster från n=-3 till n=3 med toppvärde 3 vid n=0. Figur (b)…
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Beräkna Fouriertransformen X[Ω] för de två signalerna x[n] nedan. Figur (a): ett triangelformat, symmetriskt diskret fönster från n=-3 till n=3 med toppvärde 3 vid n=0. Figur (b): triangelformat fönster förskjutet till höger (positivt n) med topp 3 vid n=3 och stöd från n=0 till n=6.
Solution (notes)
a) X[Ω] = Σ_{n=-∞}^{∞} x[n] e^{-jΩn}, där x[n] = δ[n+2] + 2δ[n+1] + 3δ[n] + 2δ[n-1] + δ[n-2] => X[Ω] = e^{j2Ω} + 2 e^{jΩ} + 3 + 2 e^{-jΩ} + e^{-j2Ω} = 3 + 4 cos(Ω) + 2 cos(2Ω) b) X[Ω] = Σ_{n=1}^{5} x[n] e^{-jΩn} där x[n] ges i a) => X[Ω] = e^{-j3Ω} (3 + 4 cos(Ω) + 2 cos(2Ω)) (förskjutning i tid motsvarar multiplikation med e^{-jΩ·3}) Kommentar: Om x_b[n] = x_a[n-3] så gäller X_b[Ω] = e^{-j3Ω} X_a[Ω].
Connections (4) est. points: 0
exam_1:4
2 p
Directly related: both tasks require computing a discrete-time frequency function from a short finite sequence of deltas and then using it to get the output to a sinusoid. Lesson 10.3 shows how to form X[Ω] from x[n]=sum of shifted deltas and how cosines appear from e^{±jΩk} terms. Relevant snippet (condensed): 'h[n] = (1/3)(δ[n] + δ[n−1] + δ[n−2]) ⇒ H(e^{jΩ}) = (1/3)(1 + e^{-jΩ} + e^{-j2Ω})'. Knowing 10.3 gives the core computation used in exam_1 Q4(c) (computing y[n] for a cosine). Estimated points: 2 (matches the subquestion size in that exam).
exam_1:6
3 p
Strong connection: exam_1 Q6(b) asks for the DFT of a finite-length discrete sequence x[n] with N0=5 (−2 ≤ n ≤ 2). Lesson 10.3 computes the DTFT/DFT of a finite sequence expressed with deltas and demonstrates the effect of index shifts (time shift → multiplication by e^{-jΩ·k}), exactly the tools needed to sketch/obtain the DFT samples. Relevant snippet (condensed): 'If x_b[n] = x_a[n−3] then X_b[Ω] = e^{-j3Ω} X_a[Ω]'. This property and the finite-sum DTFT computation map directly to the exam b) part (DFT of length 5). Estimated points: 3 (the exam part b was worth 3p).
exam_2:4
5 p
Very close connection: exam_2 Q4 requires obtaining the discrete-time spectrum X[Ω] of the sampled continuous triangular/Δ signal (resulting in a short finite discrete sequence) and then plotting |Y(Ω)| after an LTI block. Lesson 10.3 shows exactly how to compute X[Ω] for a finite combination of deltas and how a time shift multiplies the DTFT by e^{-jΩk}. Relevant snippet (condensed): 'Sampling Δ(t/2) → x[n] = 1/2 δ[n+1] + δ[n] + 1/2 δ[n−1] ⇒ X[Ω] = 1/2 e^{jΩ} + 1 + 1/2 e^{-jΩ} = 1 + cos(Ω)'. That piece (writing sampled x[n] as deltas and forming X[Ω]) gives most of the work for the exam problem; remaining steps are multiplying by H[Ω] (given). Estimated points: 5/8 (significant part but not the entire exam task).
exam_6:4
2 p
Related via core discrete-time transform techniques: exam_6 Q4 asks for the impulse response of a cascade H = H1·H2 where H1 and H2 correspond to finite-start exponential sequences. Solving can be done either by time-domain convolution of short/causal sequences or by product of transforms. Lesson 10.3 provides the basic DTFT/Z-transform intuition for finite sequences and the time-shift property (useful when aligning indices in convolution). Relevant snippet (condensed): 'h[n] = (h1 * h2)[n]; compute via Σ h1[n−m] h2[m] or via H(z)=H1(z)H2(z)'. Lesson 10.3 does not solve this specific convolution but supplies directly applicable tools (forming transforms of simple sequences and handling shifts). Estimated points: 2 (helps with parts of the solution).

10.4 0p from connections

Bestäm den inversa Fouriertransformen x[n] för de två Fouriertransformerna X[Ω] = |X[Ω]| · e^{j arg X[Ω]} nedan, specificerade i vinkelfrekvensintervallet |Ω|<π rad. (a) |X[Ω]| är…
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Bestäm den inversa Fouriertransformen x[n] för de två Fouriertransformerna X[Ω] = |X[Ω]| · e^{j arg X[Ω]} nedan, specificerade i vinkelfrekvensintervallet |Ω|<π rad. (a) |X[Ω]| är ett rektangelband: |X[Ω]| = 1 för |Ω| ≤ Ω_0, 0 annars. Arg X[Ω] är en linjär funktion som går från positiv lutning vid -Ω_0 till negativ vid +Ω_0 (skissat). (b) |X[Ω]| = rektangel som i (a). Arg X[Ω] = konstant π/2 för |Ω|≤Ω_0 och -π/2 utanför (skissat).
Solution (notes)
x[n] = (1/2π) ∫_{-π}^{π} X[Ω] e^{jΩn} dΩ a) X[Ω] = 1 för |Ω| ≤ Ω0, 0 annars => x[n] = (1/2π) ∫_{-Ω0}^{Ω0} e^{jΩ(n-n0)} dΩ = (Ω0/π) sinc( (Ω0/π)(n-n0) ) (framställning av diskret sinc) b) Uppdelning i integraler när X[Ω] ges styckvis (se handskriven lösning) => resulterande uttryck: x[n] = (1 - cos(Ω0 n)) / (π n) Kommentar i bilden visar skillnader mellan signalerna p.g.a. deras fasspektra trots lika amplitudspektra.
Connections (5) est. points: 0
exam_2 assignment 4
4 p
Snippet: "h[n] = 1/2 · sinc_N(n/2) ... Rita utsignalens amplitudspektrum |Y(Ω)|." Connection: this exam question uses the discrete ideal rectangular bandlimited frequency response ⇄ discrete-time sinc impulse-response pair. The lesson (inverse DTFT of a rectangular X[Ω] → discrete sinc x[n]) directly supplies the time-/frequency-pair and the ability to recognise/compute the sinc/rect relation and thus to form Y(Ω)=X(Ω)·H(Ω). Knowing the lesson solves the core transform step needed to draw/compute |Y(Ω)| (so gives substantial partial credit).
exam_2 assignment 2
–
Snippet: "H(ω) = e^{j2ω} sinc_N(ω/π). Beräkna y(t) för x(t)=e^{-t}u(t)." Connection: continuous-time analogue — inverse Fourier transform of a rectangular/sinc frequency shape (and the effect of the e^{j2ω} phase term as a time shift). The lesson explains the corresponding discrete inverse-FT pair (rect ⇄ sinc) and the role of phase; useful as a conceptual/prerequisite link but not identical (continuous vs discrete and additional convolution/transform steps are required).
exam_3 assignment 1
–
Snippet: "x(t)=sin(t)/t with X(ω)=π for |ω|≤1; used to compute E_y via (1/2π)∫|X(ω)H(ω)|^2 dω." Connection: uses a rectangular spectrum (continuous) and its simple time-domain counterpart (sinc). The lesson's derivation of discrete rect ⇄ discrete-sinc is directly analogous and helps understanding how rectangular spectra concentrate energy and how to manipulate such spectra (prerequisite conceptual help, but different domain so no direct points).
exam_1 assignment 6
–
Snippet: "Uniform sampling, multiply x[n] by (−1)^n, and ideal reconstruction ... Rita frekvensspektra X[Ω], Y[Ω], Y(ω)." Connection: multiplication by (−1)^n shifts the discrete spectrum by π and changes phase; the lesson explicitly comments on differences due to phase spectra despite identical amplitude spectra. Thus the lesson helps interpret phase shifts and spectral shifts (useful prerequisite), but additional sampling/aliasing/reconstruction concepts are needed for full solution.
exam_6 assignment 6
–
Snippet: "An ideal LP is preceded and followed by multiplication with (−1)^n; 'Rita det totala systemets frekvensfunktion H[Ω]'." Connection: combining modulation (multiplication by (−1)^n) with an ideal rectangular LP in frequency produces shifted rectangular passbands; the lesson's rect ⇄ sinc knowledge and understanding of phase/modulation helps predict the resulting filter type and the corresponding time-domain (sinc-like) effects. This is a strong conceptual/prerequisite connection but not identical, so no direct points assigned.

10.5 0p from connections

Utgör följande funktioner en Fouriertransform till någon tidsdiskret signal x[n]? Motivera ditt Ja/Nej. a) X[Ω] = Ω + π b) X[Ω] = j + π c) X[Ω] = sin(10 Ω) d) X[Ω] = sin(Ω/10) …
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Utgör följande funktioner en Fouriertransform till någon tidsdiskret signal x[n]? Motivera ditt Ja/Nej. a) X[Ω] = Ω + π b) X[Ω] = j + π c) X[Ω] = sin(10 Ω) d) X[Ω] = sin(Ω/10) e) X[Ω] = δ(Ω)
Solution (notes)
Bedömning om X[Ω] kan vara en Fouriertransform (måste vara 2π-periodisk) a) X[Ω] = Ω + π: Nej, inte 2π-periodisk => ingen Fouriertransform b) X[Ω] = j + π: Ja, konstant => kan vara en f-transform c) X[Ω] = sin(10Ω): Har period 2π/10 => X[Ω] = X[Ω + 2π] så kan vara Fouriertransform d) X[Ω] = sin(Ω/10): Period = 20π => inte 2π-periodisk => kan inte vara Fouriertransform e) X[Ω] = δ(Ω): Inte 2π-periodisk => kan inte vara Fouriertransform (Resonemang enligt handskrivna anteckningar.)
Connections (4) est. points: 0
exam_6 assignment 6
3 p
Snippet: "Systemet består av ett idealt lågpassfilter som föregås och efterföljs av en multiplikation med (−1)^n. a) Rita det totala systemets frekvensfunktion H[Ω]. b) Vilken typ av frekvensselektivt filter utgör det totala systemet?" (Exam_6_question.json) Why connected: The lesson is about recognizing valid DTFTs via 2π-periodicity and consequences of frequency-domain operations. Multiplication by (−1)^n = e^{jπ n} corresponds to a frequency shift Ω→Ω+π of a 2π-periodic DTFT; knowing the periodicity condition and the shift property is essential to draw the total H[Ω] and classify the resulting (shifted) filter. This directly supplies the core step in part (a) and much of (b). Estimated points: 3/8 — the lesson gives the key frequency-domain property (periodicity and shift) but additional steps (drawing the exact periodic replicas and arguing filter type) are required.
exam_2 assignment 4
3 p
Snippet: "Signalen x(t) = Δ( t / 2 ) samplas idealt med f_s = 2 Hz → x[n]. x[n] går in i ett diskret LTI-system med impulssvar h[n] = 1/2 sinc_N(n/2). Rita utsignalens amplitudspektrum |Y(Ω)|. Tip: använd inte Poisson, sampla själv." (Exam_2_question.json) Why connected: Drawing |Y(Ω)| for the sampled/discrete-time case requires knowing that the DTFT X[Ω] is 2π-periodic and how sampling and the discrete filter H[Ω] interact (multiplication in Ω). The lesson's focus on checking/using 2π-periodicity of spectra and recognizing valid DTFT forms directly helps determine the periodic repetitions and where H[Ω] applies. Estimated points: 3/8 — the periodicity/DTFT-validity is a central ingredient but one must also perform sampling-frequency scaling and the product with H[Ω].
exam_1 assignment 6
3 p
Snippet: "Rita frekvensspektren X[Ω], Y[Ω] och Y(ω) för en likformig sampling följt av multiplikation med (−1)^n och ideal rekonstruktion. Motivera grafiskt och analytiskt. Samplingsfrekvens f_s = 8 kHz." (Exam_1_question.json) Why connected: This question requires converting between continuous-time spectra and the discrete-time periodic spectrum X[Ω], and understanding how operations in time (sampling, multiplication by (−1)^n) map to shifts and periodic copies in Ω. The lesson's rule — a DTFT must be 2π-periodic and recognizing which functions are valid DTFTs — is directly used to sketch X[Ω] and its shifted copies. Estimated points: 3/8 — the lesson gives the necessary spectral-periodicity check and shift intuition, but full solution needs sampling math and reconstruction reasoning.
exam_5 assignment 6
2 p
Snippet: From a list of true/false statements: g) "Vid sampling ... kan signalen rekonstrueras om f_s ≥ 2B." h) "DFT:n till en tidsdiskret signal ... har perioden f_s/N_0 ..." — questions about sampling/DFT periodicity and frequency resolution. (Exam_5_question.json) Why connected: Several true/false items hinge on the periodicity properties of the discrete-frequency representations (DTFT/DFT) and on sampling-induced periodic replication. The lesson's focus on 2π-periodicity and recognizing valid DTFT forms gives direct justification for evaluating these statements. Estimated points: 2/8 — the lesson addresses the core periodicity concept but the exam statements also require knowledge of sampling theorem and DFT frequency-scaling.

10.6 0p from connections

Bestäm, med hjälp av formelsamlingens Tabell 7 (Fouriertransformens egenskaper) och Tabell 8 (Fouriertransformpar), Fouriertransformen X[Ω] för signalerna x[n] nedan. Antag att |a|…
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Bestäm, med hjälp av formelsamlingens Tabell 7 (Fouriertransformens egenskaper) och Tabell 8 (Fouriertransformpar), Fouriertransformen X[Ω] för signalerna x[n] nedan. Antag att |a|<1. a) x[n] = a^{n-m} u[n-m] b) x[n] = a^{n-3} (u[n] - u[n-10]) c) x[n] = a^{n-m} u[n] d) x[n] = a^{n} u[n-m]
Solution (notes)
Använd samband: Σ r^{n} u[n] ↔ e^{jΩ} / (e^{jΩ} - r), |r|<1 och tidsförskjutning x[n-n0] ↔ X[Ω] e^{-jΩ n0} a) x[n] = a^{n-m} u[n-m] = x1[n-m], där x1[n] = a^{n} u[n] => X[Ω] = e^{jΩ m} / (e^{jΩ} - a) men med anpassning enligt förskjutning: X[Ω] = e^{jΩ(1-m)} / (e^{jΩ} - a) (se bild) b) x[n] = a^{n-3} (u[n] - u[n-10]) => X[Ω] = a^{-3} e^{jΩ} / (e^{jΩ} - a) - a^{7} e^{jΩ} / (e^{jΩ} - a) e^{ - j10Ω } etc. (sammanfattning i bilden): X[Ω] = a^{-3} e^{-j3Ω} (1 - a^{10} e^{-j10Ω}) / (e^{jΩ} - a) c) x[n] = a^{n-m} u[n] => X[Ω] = a^{-m} e^{jΩ} / (e^{jΩ} - a) d) x[n] = a^{n} u[n-m] = a^{m} · a^{n-m} u[n-m] => enligt förskjutning: X[Ω] = a^{m} · (e^{jΩ(1-m)} / (e^{jΩ} - a)) (se bild för noteringar).
Connections (4) est. points: 0
exam_6 assignment 4
4 p
Very close match: the exam asks for the cascade impulse response where H1 and H2 have impulse responses that are shifted/scaled geometric sequences. Solution uses h1[n]=0.5^{n-1} u[n-1] and h2[n]=0.8^{n-1} u[n-1] and finds h[n] by convolution or via product of transforms. Lesson 10.6 gives the DTFT/Z-transform pair for Σ r^n u[n] and the time-shift property X[Ω] e^{-jΩ n0} (and formulas for a^{n-m} u[n-m]), which directly yields H1/H2 in transform domain and makes the product/inversion straightforward. Relevant snippet (condensed): "h1[n]=0.5^{n-1} u[n-1], h2[n]=0.8^{n-1} u[n-1] ⇒ h[n]=h1*h2 (time-domain sum) or H(z)=1/(z-0.5)·1/(z-0.8), invert via known transforms."
exam_4 assignment 4
4 p
Strong connection: exam gives h[n]=δ[n]+(1/3)^n u[n-1] and asks for y[n] for a given x[n]. Solving requires using the transform of a^{n} u[n] and of shifted geometric sequences (and/or using the time-shifted transform). Lesson 10.6 explicitly lists transforms for a^{n-m} u[n-m] and handling of finite windows, which directly supplies H(Ω)/H(z)=z/(z-1/3) and simplifies convolution/inversion. Relevant snippet (condensed): "h[n]=δ[n]+(1/3)^n u[n-1] ⇒ H(z)=z/(z-1/3); compute y[n]=(x*h)[n] for x[n]=u[n+3]-u[n-3] via convolution or z-domain multiplication and inversion."
exam_2 assignment 5
3 p
Related: exam asks for total impulse response of cascade where one subsystem has h2[n]=0.2^n u[n] and the other comes from a first-order difference equation (leading to 0.8^n terms). The lesson gives the transform pair for r^n u[n] and time shifts, which is a central building block for deriving the closed-form h[n]=(4·0.8^n - 0.2^n) u[n] shown in the solution. Relevant snippet (condensed): "H1: v[n]-0.8 v[n-1]=3 x[n]; H2: h2[n]=0.2^n u[n] ⇒ total h[n]=(4·0.8^n - 0.2^n) u[n]."
exam_1 assignment 5
3 p
Close/prerequisite connection: the exam computes a cascade impulse response where one stage is described by a first-order difference equation and another has h2[n]=0.2^n u[n]. Deriving/ manipulating transforms of geometric sequences and applying time shifts (as in lesson 10.6) is directly useful to obtain the total h[n]. Relevant snippet (condensed): "v[n] − 0.8 v[n−1] = 3 x[n]; h2[n]=0.2^n u[n] ⇒ compute total h[n] via Z-transforms/convolution."

10.7 0p from connections

Bestäm, med hjälp av formelsamlingens Tab. 8.5, Fouriertransformen av x[n] = a^{n} cos(Ω_0 n) u[n], där |a|<1 och 0<Ω_0<π. Kontrollera gärna sedan ditt svar med transformparet i T…
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Bestäm, med hjälp av formelsamlingens Tab. 8.5, Fouriertransformen av x[n] = a^{n} cos(Ω_0 n) u[n], där |a|<1 och 0<Ω_0<π. Kontrollera gärna sedan ditt svar med transformparet i Tab. 8:20.
Solution (notes)
x[n] = a^{n} cos(Ω0 n) u[n] = (1/2)( (a e^{jΩ0})^{n} u[n] + (a e^{-jΩ0})^{n} u[n] ) Använd tabell: X[Ω] = (1/2) · e^{jΩ} / (e^{jΩ} - a e^{jΩ0}) + (1/2) · e^{jΩ} / (e^{jΩ} - a e^{-jΩ0}) Förenkla algebraiskt enligt handskrivna beräkningar till slututtryck för X[Ω] som rationell funktion i e^{jΩ}: X[Ω] = e^{jΩ} (e^{jΩ} - a cos(Ω0)) / (e^{j2Ω} - 2 a e^{jΩ} cos(Ω0) + a^{2}) (se sista raden i bilden).
Connections (4) est. points: 0
exam_6 assignment 4
2 p
Snippet: H1[Ω]=1/(e^{jΩ}-0.5), H2 given by y[n]-0.8y[n-1]=x[n-1]; solution uses H1[z]↔h1[n]=0.5^{n-1}u[n-1], H2↔0.8^{n-1}u[n-1], then h[n]=h1*h2 computed by geometric-series algebra. Why connected: the lesson shows decomposition of a^n cos(Ω0 n) u[n] into (a e^{±jΩ0})^n u[n] and the DTFT/rational-form algebra for geometric exponentials. Those transform pairs and algebraic simplifications (expressing sums of exponentials as rational functions in e^{jΩ}) are directly used when deriving h1/h2 and when convolving two geometric sequences. The lesson thus supplies core transform/ algebra tools needed here (but the exam also requires convolution and partial-fraction/algebra beyond the lesson).
exam_1 assignment 4
1 p
Snippet: h[n]=(1/3)(δ[n]+δ[n-1]+δ[n-2]); part (c) compute y[n] for x[n]=2+5 cos(2π/3 n) using H(e^{jΩ}). Solution uses H(e^{jΩ}) and evaluates at Ω=0 and Ω=2π/3 to get DC and cos contributions. Why connected: the lesson explicitly writes cos as half-sum of complex exponentials and gives the transform of (a e^{±jΩ0})^n u[n] — the same decomposition/transform idea is needed to handle sinusoidal inputs in frequency domain and to evaluate H(e^{jΩ}) at the sinusoid frequency. It helps directly with the cos→exponential step (but does not by itself compute H(e^{jΩ}) for this specific finite impulse h[n]).
exam_4 assignment 6
1 p
Snippet: block diagram yields H(z)=(z+0.5)/(z-0.5); part (b) compute y[n] for x[n]=3·sin(π/2 n) via y[n]=3|H(e^{jΩ0})| sin(Ω0 n + arg H(e^{jΩ0})). Why connected: the lesson shows how to represent cos/sin as sums of complex exponentials and how to work in the e^{jΩ} domain (evaluate rational functions in e^{jΩ}). That representation and evaluation step is exactly what is used to turn a sinusoidal input into two complex exponentials and apply H(e^{jΩ}). The lesson supplies that elementary but necessary conversion (though the exam also needs computing H(e^{jΩ}) and extracting magnitude/phase).
exam_3 assignment 4
2 p
Snippet: difference equation y[n]+0.25 y[n-2]=2 x[n]+x[n-1]; solution derives H[z]=(2+z^{-1})/(1+0.25 z^{-2}) (a rational function in z) and states ROC. Why connected: the lesson culminates in algebraic simplification of sums of geometric exponentials to a rational function in e^{jΩ} (or z). The same algebraic manipulation and familiarity with transform pairs for sequences like a^n e^{±jΩ0 n} and with converting series sums into rational forms is directly applicable when deriving H(z) from difference equations and simplifying to rational form. It therefore supplies core techniques needed for this exam task (though solving the difference equation and ROC reasoning require additional steps).

10.8 0p from connections

Ett tidsdiskret LTI-system har frekvensfunktionen H[Ω] = (e^{jΩ} + 0.32) / (e^{j2Ω} + e^{jΩ} + 0.16) Beräkna utsignalens zero state-komponent y_{zs}[n] för de två insignalerna x[…
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Ett tidsdiskret LTI-system har frekvensfunktionen H[Ω] = (e^{jΩ} + 0.32) / (e^{j2Ω} + e^{jΩ} + 0.16) Beräkna utsignalens zero state-komponent y_{zs}[n] för de två insignalerna x[n] nedan. a) x[n] = (-0.5)^{n} · u[n] b) x[n] = u[n]
Solution (notes)
LTI-system: y_{2s}[n] = (x * h)[n] ⇔ Y_{2s}[Ω] = X[Ω] · H[Ω] a) X[Ω] = 2·(0.5)^{n} u[n] ↔ e^{jΩ} / (e^{jΩ} + 0.5) enligt tabell. H[Ω] = (e^{jΩ} + 0.32)/(e^{j2Ω} + e^{jΩ} + 0.16) enligt uppgift. => Y_{2s}[Ω] = (e^{jΩ} / (e^{jΩ} + 0.5)) · ((e^{jΩ} + 0.32)/(e^{j2Ω} + e^{jΩ} + 0.16)) Partialbråksuppdelning ger i tidsdomänen (enligt Tab. 8:5) lösning: y_{2s}[n] = (2 (−0.5)^{n} − (8/3)(−0.8)^{n} + (2/3)(−0.2)^{n}) u[n] Kommentar: Lösningen visar samma typ av beräkningar som i z-transformen; ofta enklare att hantera z-transformer. b) X[Ω] = u[n] => enligt Tab. 8:3 X[Ω] = vp{ e^{jΩ} / (e^{jΩ} − 1) } + π Σ δ(Ω − k·2π) Genom att betrakta intervallet −π ≤ Ω ≤ π och använda principalvärde samt delta-termer fås partialbråksuppdelning och tidsdomänsvar: y_{2s}[n] = (132/216)·{vp{ e^{jΩ} / (e^{jΩ} − 1) } + π Σ δ(Ω − k·2π)} + … => i tidsdomänen y_{2s}[n] = (132/216) u[n] − (4/9)(−0.8)^{n} − (1/6)(−0.2)^{n} u[n] Kommentar: Uppgift b) visar att Fouriertransformberäkningar kan bli omfattande; z-transform ofta enklare.
Connections (9) est. points: 0
exam_6 - 4
6 p
Snippet: “H1[Ω] = 1/(e^{jΩ} − 0.5). H2 via y[n] − 0.8 y[n−1] = x[n−1]. Beräkna totala kaskadkopplade systemets impulssvar h[n] genom a) tidsdomänen b) transformdomänen.” (lösning ger h[n] as sum of exponentials). Connection: This is a direct application of discrete-time LTI theory Y[Ω]=X[Ω]·H[Ω] and convolution/Z-domain multiplication; the exam explicitly asks for h[n] by both time-domain convolution and transform-domain algebra/partial fractions — the same techniques as in lesson 10.8. If a student mastered 10.8 they would solve most of this question; remaining bookkeeping (ROC, index shifts) account for the gap, hence 6 pts.
exam_1 - 5
5 p
Snippet: “H1: v[n] − 0.8 v[n−1] = 3 x[n]. H2: h2[n] = 0.2^n u[n]. a) Beräkna totala kaskadkopplade systemets impulssvar h[n].” (lösning uses Z-transform products and inverse partial-fraction expansion). Connection: computing cascade impulse response by convolving exponentials / multiplying transfer functions in z/DTFT domain and doing partial-fraction inversion — core content of 10.8. Several extra steps (difference-equation manipulation, block realization) explain not giving full 8 pts.
exam_2 - 5
4 p
Snippet: “Givet x[n]=(0.5^n + 1) u[n] och y[n]=2δ[n] − 1.5δ[n−1]. a) Beräkna systemfunktionen H(z) … b) Skissera |H(Ω)|.” Connection: This is the inverse problem X·H = Y in transform domain; lesson 10.8 teaches forming X[Ω], H[Ω] and using algebra/partial fractions to get time-domain responses. Knowing 10.8 gives the main method to compute H(z)/H(Ω), though some algebraic rearrangement and ROC reasoning remain, so 4 pts.
exam_4 - 4
5 p
Snippet: “h[n] = δ[n] + (1/3)^n u[n−1]. b) Beräkna y[n] för x[n] = u[n+3] − u[n−3].” Connection: Direct convolution of a given finite/causal impulse response with a piecewise step-like x[n]; lesson 10.8 shows forming X[Ω], H[Ω], multiplying and inverting (or doing time-domain convolution). Mastery of the lesson reduces this to routine convolution / transform inversion; some casewise indexing work remains, so 5 pts.
exam_2 - 6
4 p
Snippet: (signal-flow with delays) “a) Beräkna systemets systemfunktion H(z). b) Beräkna y[n] för x[n]=cos(π/3 n).” (solution computes H(z), evaluates H(e^{jΩ0}) and forms y[n]=|H| cos(Ω0 n + arg H)). Connection: Lesson 10.8 emphasises Y[Ω]=X[Ω]·H[Ω] and using H(e^{jΩ}) to scale/phase sinusoids — directly used in part b. Knowing 10.8 gives the main step (evaluate H on unit circle) — hence 4 pts.
exam_3 - 4
3 p
Snippet: “y[n] + 0.25 y[n−2] = 2 x[n] + x[n−1]. a) Beräkna systemfunktionen H[z] … b) Beräkna zero-input response y_zi[n].” Connection: Uses z-transform to get H(z) and partial-fraction inversion to obtain homogeneous solution (exponentials multiplied by polynomials in n) — techniques highlighted in 10.8 (and the solution comment that z-transform often easier). Knowing 10.8 gives a strong basis; additional initial-condition handling reduces credit to 3 pts.
exam_1 - 4
4 p
Snippet: “h[n] = (1/3)(δ[n] + δ[n−1] + δ[n−2]). b) Skissera |H(Ω)| … c) Beräkna y[n] för x[n] = 2 + 5 cos(2π/3 n).” Connection: Computing H(e^{jΩ}) from finite h[n] and obtaining output for a sum of DC + sinusoid via multiplication in frequency (or direct convolution) is a core application of 10.8. A student who knows 10.8 can perform these steps; small work on plotting/interpretation accounts for partial points.
exam_6 - 6
3 p
Snippet: “System: multiplication by (−1)^n before and after an ideal LP. a) Rita totala H[Ω]. b) Vilken typ av filter utgör det totala systemet?” Connection: This uses the modulation/ frequency-shift property in discrete time and composing H(Ω) via multiplication/shift — related to the frequency-domain manipulations emphasized in lesson 10.8 (forming H and applying to signals). It's less about partial fractions but directly about working with H[Ω], so counted with 3 pts (motivation-only style connection).
exam_2 - 4
3 p
Snippet: “x(t)=Δ(t/2) sampled → x[n] → LTI h[n]=1/2 sinc_N(n/2). Rita |Y(Ω)|. Tip: don't use Poisson; sample directly.” Connection: While this is sampling/continuous→discrete, the final step is computing Y[Ω] = X[Ω]·H[Ω] (product of discrete spectra) — exactly the relation from 10.8. A student knowing 10.8 would form the product and sketch |Y(Ω)|; remaining sampling specifics account for reduced points.

10.9 0p from connections

Ett ackumulatorsystem har den egenskapen att det, för insignalen x[n], genererar utsignalen y[n] = Σ_{k=-∞}^{n} x[k]. a) Är detta ett tidsdiskret LTI-system? Motivera! b) Beräkna…
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Ett ackumulatorsystem har den egenskapen att det, för insignalen x[n], genererar utsignalen y[n] = Σ_{k=-∞}^{n} x[k]. a) Är detta ett tidsdiskret LTI-system? Motivera! b) Beräkna systemets impulsvar h[n] och frekvensfunktion H[Ω].
Solution (notes)
a) Förhållandet mellan utsignal y[n] och insignal x[n] är en differensekvation med konstanta koefficienter, vilket innebär att systemet är linjärt och tidsinvariant (LTI). Exempelvis y[n] = Σ_{k=-∞}^{n} x[k] = x[n] + x[n-1] + x[n-2] + … b) Låt x[n] = δ[n] => y[n] = h2[n] = Σ_{k=-∞}^{∞} δ[n-k] = δ[n] + δ[n-1] + δ[n-2] + … = u[n] Formelsamlingen Tab. 8:3 ger H[Ω] för detta system: H[Ω] = vp{ e^{jΩ} / (e^{jΩ} − 1) } + π Σ_{k=-∞}^{∞} δ(Ω − k·2π). (Se handskrift för motiv och resonemang.)
Connections (6) est. points: 0
exam_1:4
4 p
Directly related: exam asks for the step response g[n] given an impulse response h[n]. The lesson shows the summation relation between output and input y[n] = Σ_{k=-∞}^{n} x[k] and that the impulse response to δ[n] is h[n]=u[n] (i.e. the running sum). Knowledge that the step response equals the cumulative sum of h (g[n]=Σ_{k=-∞}^n h[k]) and the role of u[n] makes the exam subproblem much easier. Snippet: "Bestäm systemets stegsvar g[n]." (solution uses g[n]=Σ h[k] / expresses g via shifted u[n]).
exam_3:5
4 p
Strong connection: the exam (Q5) asks for the system step response g[n] via z-transform or convolution and the solution expresses g[n] as sums of shifted unit-steps u[n−k] and piecewise integer values (n+1, etc.). The lesson teaches that the summation operator has impulse response u[n] and that convolution with u (or summing) produces cumulative sums — directly useful to derive the piecewise form. Snippet: "Beräkna systemets stegsvar g[n] ... resultaat g[n]=0 for n<0, g[n]=n+1 for 0≤n≤5, g[n]=6 for n≥6."
exam_6:4
5 p
Very relevant: H1[Ω]=1/(e^{jΩ} − 0.5) is the frequency-domain object whose inverse gives a causal exponential times a step (similar to the table entry for 1/(1 − e^{-jΩ}) ↔ u[n]). The lesson's DTFT formula for the unit-step (H[Ω] = vp{ e^{jΩ}/(e^{jΩ} − 1) } + π Σ δ(Ω−2πk)) and attendant mapping between simple rational e^{jΩ}-forms and time-domain u[n]-type sequences helps directly to invert H1 and compute the cascade impulse response by convolution. Snippet: "H1[Ω] = 1/(e^{jΩ} − 0,5) ... beräkna det totala kaskadkopplade systemets impulssvar h[n] ... lösning: h[n]=(25/6·0.8^n − 20/3·0.5^n)u[n−1] (eller equivalent)."
exam_1:5
3 p
Related: exam asks to cascade a system given by a difference equation (v[n]−0.8v[n−1]=3x[n]) with another having h2[n]=0.2^n u[n], and compute the total impulse response h[n]. The lesson's discussion of impulse response equal to u[n] (and use of u[n] in convolution) helps understanding how to combine/interpret IIR impulse responses and their convolution with unit-step-like sequences; useful background for computing h[n] by convolution/Z-transform. Snippet: "H1 differensekvation v[n] − 0,8 v[n−1] = 3 x[n]. System H2 impulssvar h2[n] = 0,2^n u[n]. Beräkna totala impulssvaret h[n]."
exam_4:4
3 p
Relevant: this question gives h[n] = δ[n] + (1/3)^n u[n−1] and asks about causality/stability and to compute y[n] for an input composed of shifted steps (u[n+3] − u[n−3]). The lesson's emphasis on u[n] as impulse response of the summation operator and on manipulating sums of shifted unit-steps directly supports the time-domain convolution/ summation techniques used in the solution. Snippet: "h[n] = δ[n] + (1/3)^n · u[n−1]. Beräkna systemets utsignal y[n] för x[n] = u[n+3] − u[n−3]."
exam_5:4
2 p
Related/prerequisite: very similar theme to exam_4:4 (same type of h[n] with a u[n−k] term). The lesson's presentation of u[n] and its transform with impulses in frequency helps reasoning about causality/stability and about effects of adding step-like terms to h[n], but it does not by itself compute the concrete convolution required in the exam. Snippet: "h[n] = δ[n] + (1/3)^n u[n−1] — a) Rita pol-nollställa; b) kausalitet; c) skissera |H(Ω)|; e) beräkna impulssvar h[n]."

10.10 0p from connections

Ett tidsdiskret energifritt LTI-system har frekvensfunktionen H[Ω], sådan att det i intervallet |Ω|<π kan uttryckas som Ĥ[Ω] = rect(Ω / π) · e^{jΩ n_0}. (Notationen i bilden visar …
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Ett tidsdiskret energifritt LTI-system har frekvensfunktionen H[Ω], sådan att det i intervallet |Ω|<π kan uttryckas som Ĥ[Ω] = rect(Ω / π) · e^{jΩ n_0}. (Notationen i bilden visar rect(Ω / 2Ω_c) · e^{j2Ω n_0} men här är formuleringen: Ĥ[Ω] = rect((Ω)/(π)) · e^{jΩ n_0}.) Beräkna systemets utsignal y[n] för de olika insignalerna x[n] nedan. a) x[n] = sinc_c(n / 2) b) x[n] = sinc_c(n) c) x[n] = sinc_c^{2} (n / 4) (Anm: uppgiften i originalbilden visar exempel med ideal lågpass och olika skalade sinc-signaler.)
Solution (notes)
Energifritt LTI-system innebär y[n] = y_{2s}[n] = (x * h)[n] ⇔ Y[Ω] = X[Ω]·H[Ω] I intervallet |Ω|<π kan X[Ω], H[Ω] uttryckas som X̂[Ω], Ĥ[Ω] och Ŷ[Ω] = X̂[Ω] Ĥ[Ω]. a) x[n] = sinc_N (n/2) (diskret sinc med Ω0 = π/2) => X̂[Ω] = (π/π/2) · rect(Ω/π) = 2·rect(Ω/π) => Ŷ[Ω] = X̂[Ω] Ĥ[Ω] = 2·rect(Ω/π)·rect(Ω/π)·e^{-j2Ω} = 2·rect(Ω/π)·e^{-j2Ω} => y[n] = sinc_N ( (n-2)/2 ) (enligt Tab. 7:6 & 8:10) b) x[n] = sinc_N(n) = sin(π n)/ (π n) => X[Ω] = 1 => Ŷ[Ω] = Ĥ[Ω] => y[n] = (1/2) sinc_N( (n-2)/2 ) (enligt tabellerna i noterna) c) x[n] = sinc_N^{2}(n/2) => med Ω0/2π = 1/4 (dvs Ω0 = π/2) och tabeller ger slutresultat y[n] = sinc_N^{2}( (n-2)/4 ) (se bild för detaljer). (Se handskrivna steg och hänvisningar till formler i Tab. 7 och 8.)
Connections (4) est. points: 0
exam_2_q4
4 p
Snippel: "x(t)=Δ(t/2) samplas med f_s=2 Hz → x[n]; systemets impulssvar h[n] = 1/2 · sinc_N(n/2). Rita |Y(Ω)|. Tips: använd inte Poisson, sampla själv.". Varför kopplat: Uppgiften kräver exakt samma kärnidé som i lektionen: H[Ω] för h[n]= (1/2)·sinc_N(n/2) är en rektangulär passband (ideal bandbegränsning ±π/2) och Y[Ω]=X[Ω]·H[Ω]. Om man kan tabellkopplingen rect↔sinc och multiplikation i frekvensdomänen (såsom i lektionen) får man direkt formen på |Y(Ω)|; det återstår att bestämma X[Ω] från sampling av Δ(t/2) (ytterligare steg). Därför ger lektionen en stor del av lösningen men inte alla detaljer (sampling/continuous→discrete conversion), uppskattade poäng: 4/8.
exam_6_q4
3 p
Snippel: "H1[Ω] = 1/(e^{jΩ}-0.5) (ger h1[n]=0.5^{n-1}u[n-1]) och H2 ges via y[n]-0.8y[n-1]=x[n-1]; kaskadkoppla H1 och H2 och beräkna totala impulssvaret h[n]—både i tidsdomänen och i transformdomänen.". Varför kopplat: Lektionen visar principen 'energifritt LTI: y[n]=(x*h)[n] ⇔ Y[Ω]=X[Ω]·H[Ω]' och hur tabellpar för exponentiella sekvenser respektive rekt/ sinc används för att gå mellan tids- och frekvensdomäner. I den här tenta­uppgiften används exakt samma convolution⇔multiplikationstanke (kaskad → produkten av överföringsfunktioner i frekvensområdet) och inversion till tidsdomänen med geometriska serier. Lektionen ger den centrala metodiken (multiplicera H1·H2 i frekvens eller konvolvera h1*h2 i tidsdomänen); kvarstående arbete är algebra/partiella summor. Uppskattade poäng: 3/8.
exam_1_q6
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Snippel: "Kaskad: likformig sampling → multiplikation av x[n] med (−1)^n → ideal rekonstruktion (PAM). f_s=8 kHz. Insignal x(t) är en sinc^2 med triangel­formad spektralgraf. a) Rita X[Ω], Y[Ω] och Y(ω).". Varför kopplat (utan poäng): Lektionen behandlar DTFT, rekt/sinc-tabeller och hur multiplicering i frekvens ger utsignalens spektrum (Y= X·H). Här uppträdde liknande byggstenar (diskret spektrum, idealfilteregenskaper) men uppgiften kräver ytterligare koncept: sampling (continuous→discrete mapping), tidsdomänsförskjutningar och tidsdomänsmultiplikation med (−1)^n (modulation/shift i frekvens), samt återuppbyggnad via PAM. Kunskap från lektionen är direkt användbar som grund men ensam räcker inte för hela lösningen; därför inga direkta poäng.
exam_6_q6
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Snippel: "System: multiplikation med (−1)^n före och efter ett idealt lågpassfilter H_LP[Ω]. a) Rita totala H[Ω]. b) Vilken typ av filter utgör det totala systemet?". Varför kopplat (utan poäng): Lektionen visar hur frekvensfunktioner multipliceras och hur en fas­faktor e^{-jΩn0} motsvarar tidsförskjutning. Uppgiften här använder relaterade DTFT-egenskaper (multiplikation i tidsdomänen ↔ spektralförskjutning, och produkt av frekvensfunktioner när man kaskadkopplar). Men den viktigaste operationen i tentan är modulation med (−1)^n som skiftar spektrum med π — ett begrepp som inte var huvudfokus i lektionen. Därför är kopplingen konceptuell/prerequisite men inte direkt poänggivande.

10.11 0p from connections

Det ideala lågpassfiltret nedan har impulssvaret h[n] och frekvensfunktionen H[Ω] = Σ_{k=-∞}^{∞} rect((Ω - k·2π) / 2Ω_c), dvs. H[Ω] = Ĥ[Ω] = rect(Ω / 2Ω_c) i intervallet |Ω|<π ra…
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Det ideala lågpassfiltret nedan har impulssvaret h[n] och frekvensfunktionen H[Ω] = Σ_{k=-∞}^{∞} rect((Ω - k·2π) / 2Ω_c), dvs. H[Ω] = Ĥ[Ω] = rect(Ω / 2Ω_c) i intervallet |Ω|<π rad och är för övrigt 2π-periodisk. (a) Ett annat LTI-system har impulssvaret h_2[n] = (-1)^{n} · h[n]. - Beräkna frekvensfunktionen H_2[Ω] för detta system. - Rita även H_2[Ω] i intervallet -4π ≤ Ω ≤ 4π rad och ange, med motivering, vilken typ av frekvensselektivt filter detta är. (b) Systemet nedan är en realisering av systemet i uppgift (a), dvs. det inre systemet är lågpassfiltret med impulsvarande h[n] och frekvensfunktionen H[Ω] enligt ovan. Det totala systemet, som är streckat, har impulsvar h_2[n] och frekvensfunktionen H_2[Ω]. (I figuren: x[n] multipliceras med (-1)^n innan och efter ett idealt LP-block; visa, utgående från denna realisering, att h_2[n] = (-1)^n h[n]. Förklara även hur filtreringen av x[n] praktiskt går till här.)
Solution (notes)
a) H2[Ω] = Σ_{k=-∞}^{∞} rect( (Ω − k·2π) / (2 Ω_c) ) Formelsamlingens Tab. 8:10 ger h2[n] = (Ω_c/π) sinc_N( (Ω_c/π) n ). Vidare gäller h2[n] = (−1)^{n} h1[n] ⇒ H2[Ω] = H1[Ω − π] = Σ rect( (Ω − (2k+1)π) / (2 Ω_c) ). Diagram i bilden visar att detta system spärra tidsdiskreta frekvenssignaler för |Ω|<Ω_c medan frekvenssignaler nära Ω = ±π passerar. Detta är ett idealiskt högpassfilter. b) Inför hjälpsignalerna g[n] och r[n] enligt figuren: inre systemet är ett ideal LP-filter medan yttre systemet är ett idealt HP-filter. Genom algebraisk manipulation och tids/frekvensförskjutningar visas att totala systemet är ett HP-filter och att h2[n] = (−1)^{n} h1[n], samt i frekvensdomänen H2[Ω] = H1[Ω − π]. Kommentar: Multiplikation med (−1)^{n} förskjuter spektrum i frekvensdomänen från att ligga runt Ω=0 till att ligga runt Ω=±π, vilket är anledningen till HP-egenskaperna.
Connections (3) est. points: 0
exam_6 assignment 6
8 p
Snippet: "The discrete system consists of an ideal lowpass (HLP[Ω]) with a multiplication by (−1)^n before and after the LP; a) draw total H[Ω]; b) what type of frequency-selective filter is the total system?" — Solution uses: multiplication by (−1)^n shifts the DT spectrum by π so the total system is the LP shifted to ±π (i.e. an ideal highpass). Connection: IDENTICAL concept. The lesson shows h2[n]= (−1)^n h1[n], H2[Ω]=H1[Ω−π] and explains that multiplying by (−1)^n shifts energy from Ω≈0 to Ω≈±π, producing highpass behaviour. This directly gives the full solution to the exam question.
exam_1 assignment 6
3 p
Snippet: "Cascade: uniform sampling → multiply sampled signal x[n] by (−1)^n → ideal PAM reconstruction. a) Draw X[Ω], Y[Ω] and Y(ω)." — Connection: core step is the effect of multiplication by (−1)^n (DT frequency shift by π) on the sampled-signal spectrum; the lesson explains this property and that it turns an LP-type spectrum around Ω=0 into components near ±π (i.e. high-frequency DT components). Additional topics required by the exam (sampling/aliasing, DT↔CT frequency scaling and PAM reconstruction) are outside the single lesson, so only partial credit is awarded (estimate 3/8 points).
exam_2 assignment 4
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Snippet: "A continuous signal is sampled and then passed to a discrete-time LTI system with h[n] = (1/2) sinc_N(n/2) (an ideal DT lowpass). Draw the output amplitude spectrum |Y(Ω)|." — Connection (prerequisite): the lesson treats ideal DT rectangular spectra and manipulation of ideal LP/HP responses in the DT domain and explains how spectral support maps under operations (shifts/multiplications). That knowledge helps when sketching |Y(Ω)| after sampling and DT filtering, but the exam question centers on sampling and DT→CT mapping rather than the (−1)^n modulation result from the lesson, so no direct points are assigned (prerequisite/motivation only).