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Digital Signal Processing End Term: 13 April 2025 (January 2025 term)

Question 1

+50 marksOne correct option
  1. Consider a stable LTI system whose system function is given below

H(z)=(1−4z−2)(1+12z−1)1−12z−1H(z) = \frac{(1 - 4z^{-2})\left(1 + \frac{1}{2}z^{-1}\right)}{1 - \frac{1}{2}z^{-1}}

(a) Determine the poles and zeros of the system. Is it a minimum phase system? [2 Marks]

(b) The system function H(z)H(z) can be represented as a cascade of a minimum – phase system Hmin(z)H_{min}(z) and a unity gain all pass system Hap(z)H_{ap}(z). Determine a choice for Hmin(z)H_{min}(z) and Hap(z)H_{ap}(z). [3 Marks]

(c) Is the minimum phase system Hmin(z)H_{min}(z) obtained in part (b) an FIR system? Justify. [2 Marks]

(d) Is the minimum – phase system Hmin(z)H_{min}(z) obtained in part (b) a linear phase system? If not can you represent H(z)H(z) as a cascade of a linear phase system, Hlin(z)H_{lin}(z) and a unity gain all – pass system Hap1(z)H_{ap1}(z)? [3 Marks]

  1. Consider a Linear Time Invariant system having an impulse response

h[n]=δ[n]−3δ[n−1]h[n] = \delta[n] - 3\delta[n-1]

Suppose the following input is given to the system

x[n]=2δ[n]+3δ[n−1]+4δ[n−2]x[n] = 2\delta[n] + 3\delta[n-1] + 4\delta[n-2]

Using DFT, find the output of the system.

  1. Consider a discrete time causal LTI FIR filter whose impulse response h[n]h[n] is non zero only over five consecutive time samples. The frequency response of the filter is H(ejω)H\left(e^{j\omega}\right).

The following information is given about the filter

a. ∫−ππH(ejω)dω=π\int_{-\pi}^{\pi} H\left(e^{j\omega}\right) d\omega = \pi

b. The frequency response can be written as

H(ejω)=A(ω)e−j2ωH\left(e^{j\omega}\right) = A(\omega)e^{-j2\omega}

Here A(ω)A(\omega) is real and even.

c. A(0)=1A(0) = 1 and A(π)=3A(\pi) = 3

i. Justify that it is a Type – I filter. [2 Marks]
ii. Completely specify the impulse response h[n]h[n] and plot it. [5 Marks]
iii. Find the frequencies which this filter will completely block or stop. [3 Marks]

  1. Consider a length 14 discrete time sequence defined for 0≤n≤130 \leq n \leq 13,
nn012345678910111213
x[n]x[n]3−1-1203−2-20aa−4-46bb345

Let the 14-point DFT be denoted as X[k],0≤k≤13X[k], 0 \leq k \leq 13.
Given that X[0]=22X[0] = 22 and X[7]=−2X[7] = -2.

Evaluate the following without computing the DFT

(i) The missing samples, aa and bb [3 Marks]
(ii) ∑k=013X[k]\sum_{k=0}^{13} X[k] [2 Marks]
(iii) ∑k=013e−j5π7k X[k]\sum_{k=0}^{13} e^{-j\frac{5\pi}{7}k}\, X[k] [3 Marks]
(iv) ∑k=013∣X[k]∣2\sum_{k=0}^{13} |X[k]|^2 [2 Marks]

  1. A

    I have written answers on the answer sheets

  2. B

    Not applicable

More on the Digital Signal Processing End Term 13 Apr 2025 paper

The IIT Madras BS Digital Signal Processing (Digital Signal Processing) End Term paper sat on 13 Apr 2025, in the January 2025 term: 1 question for 50 marks in 180 minutes. The first 1 question is below. Sign in with Google — it is free — to see the whole paper with its answers and explanations, in learning mode or as a timed mock test.

FeatureDigital Signal Processing End Term 13 Apr 2025 at a glance
TermJanuary 2025 term
SubjectDigital Signal Processing
Course codeEE3101
Questions1
Marks50
Duration180 min
MCQ1
Official paperIIT M ES FOUNDATION FN EXAM QEF1 13 Apr 2025
Negative markingNo negative marking.
Updated

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