Compiled DSP Questions: Answers 51–75 (ADC, Sampling, Z-Transform, DFT and FFT)

Worked answers to questions 51-75 of the compiled CSE-409 DSP question list, with sampling numericals, inverse Z-transform and 8-point DFT.

Posted Mar 20, 2026

By Sadman Ishtiak

views 9 min read

How This Post Fits the Series

Questions 51–75 are taken in order from the Compiled DSP Questions for CSE-409 page. Full derivations live in the linked posts; this page is the revision sheet.

Q51. What is a signal and how is it converted to digital?

A signal is a quantity that varies with time or space and carries information, described mathematically as $x(t)$ or $x[n]$. To digitize it: condition and anti-alias filter the analog voltage, sample it at $F_s\ge2F_{\max}$, quantize each sample to $2^B$ levels, then encode the levels as binary words. The output is a sequence of code words.

Q52. DSP system block diagram

x_a(t) -> [Anti-alias LPF] -> [Sample & Hold] -> [ADC: quantize + encode]
       -> [Digital processor: filter, FFT, detect] -> [DAC]
       -> [Reconstruction LPF] -> y_a(t)

The anti-alias filter removes content above $F_s/2$ before sampling. The ADC makes the signal digital. The processor performs the actual DSP algorithm. The DAC and reconstruction filter return the result to continuous time.

Q53. State the sampling theorem and define the Nyquist rate

Sampling theorem: a band-limited signal with highest frequency $F_{\max}$ can be reconstructed exactly from its samples if the sampling frequency satisfies

\[F_s>2F_{\max}\]

The Nyquist rate is $2F_{\max}$, the minimum sampling rate that avoids aliasing. The Nyquist frequency is $F_s/2$, the highest frequency the sampler can represent at a given $F_s$.

Q54. Graphical representations of $\delta[n]$ and $u[n]$

The unit impulse has a single stem of height 1 at $n=0$ and zero elsewhere:

$n$	$-2$	$-1$	$0$	$1$	$2$
$\delta[n]$	$0$	$0$	$1$	$0$	$0$

The unit step has stems of height 1 for every $n\ge0$:

$n$	$-2$	$-1$	$0$	$1$	$2$
$u[n]$	$0$	$0$	$1$	$1$	$1$

Graphically these are stem plots, not continuous curves. Note $\delta[n]=u[n]-u[n-1]$.

Q55. Continuous-time versus discrete-time signals

Continuous-time signals are defined for every real $t$ and are described by $x(t)$; their amplitude may take any value. Discrete-time signals are defined only at integer indices $n$ and are described by $x[n]$. A discrete-time signal may still have continuous amplitude, as after sampling but before quantization. Discrete-time signals are usually obtained by sampling, $x[n]=x_a(nT_s)$, and are the natural input to a computer.

Q56. A 6 Hz sinusoid for 0–2 s then a 4 Hz sinusoid for 2–4 s: is it stationary?

The signal is

\[x(t)=\begin{cases}\sin(12\pi t),&0\le t<2\\ \sin(8\pi t),&2\le t<4\end{cases}\]

Sketch: two full 6 Hz cycles in the first two seconds are followed by two full 4 Hz cycles in the next two seconds. The waveform changes frequency abruptly at $t=2$ s.

It is not stationary. A stationary signal has statistical properties, including frequency content, that do not change with time. Here the spectrum and the zero-crossing rate change at $t=2$ s, so the signal is non-stationary (piecewise stationary at best).

Q57. Procedure to transform a continuous signal to digital

Transduce the physical quantity to a voltage, amplify and condition it, low-pass filter it to prevent aliasing, sample and hold at $F_s$, quantize to $2^B$ levels, and encode each level into a binary word. Optionally compress or packetize the resulting bit stream.

Q58. Define aliasing and how to avoid it

Aliasing is the overlap of spectral replicas that occurs when $F_s<2F_{\max}$. A frequency $F>F_s/2$ appears at the lower frequency $F_s-F$, so the reconstructed signal is wrong and the error is irreversible.

Avoid it by (1) choosing $F_s\ge2F_{\max}$, (2) placing an anti-alias low-pass filter before the sampler, and (3) oversampling so the analog filter has a wide transition band.

Q59. $x_a(t)=3\cos 2000\pi t+5\sin 6000\pi t+10\cos 12000\pi t$

The frequencies are $F=1000$, $3000$, and $6000$ Hz. The maximum is $6000$ Hz.

a) Nyquist rate.

\[F_s^{\min}=2F_{\max}=12000\ \text{Hz}\]

b) $F_s=5000$ samples/s.

\[x[n]=3\cos\left(\frac{2\pi\cdot1000}{5000}n\right)+5\sin\left(\frac{2\pi\cdot3000}{5000}n\right)+10\cos\left(\frac{2\pi\cdot6000}{5000}n\right)\] \[x[n]=3\cos(0.4\pi n)+5\sin(1.2\pi n)+10\cos(2.4\pi n)\]

Folding with $\sin(1.2\pi n)=-\sin(0.8\pi n)$ and $\cos(2.4\pi n)=\cos(0.4\pi n)$:

\[x[n]=13\cos(0.4\pi n)-5\sin(0.8\pi n)\]

The 3000 Hz component aliases to 2000 Hz and the 6000 Hz component aliases to 1000 Hz, adding to the original 1000 Hz term.

Q60. Define skewness and kurtosis

Skewness is the normalized third central moment:

\[\text{skew}=\frac{1}{N}\sum\left(\frac{x_i-\mu}{\sigma}\right)^3\]

It measures asymmetry: zero for a symmetric distribution, positive for a long right tail, negative for a long left tail.

Kurtosis is the normalized fourth central moment:

\[\text{kurt}=\frac{1}{N}\sum\left(\frac{x_i-\mu}{\sigma}\right)^4\]

It measures tail heaviness and peakedness. A normal distribution has kurtosis 3; values above 3 are heavy-tailed (leptokurtic), below 3 are light-tailed (platykurtic). Excess kurtosis is kurtosis minus 3.

Q61. Why quantization is the “opposite” of sampling

Sampling discretizes the time axis but leaves the amplitude continuous. Quantization discretizes the amplitude axis but leaves time continuous. Each acts on the axis the other leaves alone; together they turn an analog signal into a digital one. Sampling error is aliasing (removable only by filtering before sampling), while quantization error is irreversible rounding noise added after sampling.

Q62. Time shifting and finding $x(n-3)$, $x(n+2)$

Time shifting replaces the index: $x[n-k]$ delays the signal by $k$ samples (moves it right), and $x[n+k]$ advances it by $k$ samples (moves it left). If the original non-zero support is $n_1\le n\le n_2$, then $x[n-3]$ has support $n_1+3\le n\le n_2+3$ and $x[n+2]$ has support $n_1-2\le n\le n_2-2$. For a sketch, copy every stem and move it three places right for $x[n-3]$, two places left for $x[n+2]$.

Q63. Differences between the Z-transform and the Fourier transform

Property	Z-transform	Fourier transform
Variable	complex $z=re^{j\omega}$	real frequency $\omega$ or $F$
Scope	wider; handles growing and decaying signals through $r$	bounded, steady-state signals
Relation	Fourier transform is $X(z)$ on the unit circle $z=e^{j\omega}$	special case of the Z-transform
ROC	explicitly required	convergence implied by integrability
Use	difference equations, stability, transients	spectral content, frequency response

Q64. Z-transform and ROC of $x[n]=a^n u[n]$

\[X(z)=\sum_{n=0}^{\infty}a^n z^{-n}=\sum_{n=0}^{\infty}(az^{-1})^n=\frac{1}{1-az^{-1}}=\frac{z}{z-a}\]

The geometric series converges when $\vert az^{-1}\vert<1$, i.e.

\[\text{ROC: }\vert z\vert>\vert a\vert\]

The pole is at $z=a$ and the ROC is outside it, as expected for a right-sided sequence.

Q65. Inverse Z-transform of $X(z)=\dfrac{z+0.2}{(z+0.5)(z-1)}$, $\vert z\vert>1$

The ROC is outside the unit circle, so the sequence is causal. Expanding $X(z)/z$ and using the residues already computed in Q41:

\[X(z)=-0.4-\frac{0.4z}{z+0.5}+\frac{0.8z}{z-1}\] \[x[n]=-0.4\delta[n]-0.4(-0.5)^n u[n]+0.8u[n]\]

Equivalently $x[n]=\left(0.8-0.4(-0.5)^n\right)u[n]-0.4\delta[n]$.

Q66. What is FFT?

The fast Fourier transform is a family of algorithms that computes the DFT with far fewer operations than the direct sum. Radix-2 FFT reduces the cost from $N^2$ to $(N/2)\log_2 N$ complex multiplications by recursively splitting the sequence into even and odd halves. The output is exactly the DFT; only the computation changes.

Q67. Fourier transform of $x(t)=e^{-a\vert t\vert}$, $a>0$

Split the integral at zero:

\[X(\omega)=\int_{-\infty}^{0}e^{at}e^{-j\omega t}\,dt+\int_{0}^{\infty}e^{-at}e^{-j\omega t}\,dt\] \[X(\omega)=\frac{1}{a-j\omega}+\frac{1}{a+j\omega}=\frac{2a}{a^2+\omega^2}\]

This is a real, even, positive spectrum with a peak of $2/a$ at $\omega=0$; the signal is a two-sided exponential, so its transform is a Lorentzian.

Q68. DFT of ${1,1,0,0}$ and IDFT of ${1,0,1,0}$

These are the same pair as Q44.

DFT: $X={2,\,1-j,\,0,\,1+j}$.

IDFT: $x={0.5,\,0,\,0.5,\,0}$.

Q69. Explaining the FFT speed factor of 21.33

For $N=64$, the direct DFT needs $64^2=4096$ complex multiplications. The radix-2 FFT needs

\[\frac{N}{2}\log_2 N=32\times6=192\]

multiplications. The ratio is $4096/192=21.33$. In general the ratio is $2N/\log_2 N$, so the advantage grows with transform length. This is why FFT-based spectrum analysis, OFDM, and image processing are practical.

Q70. Radix-2 DIT-FFT algorithm

Reorder the input in bit-reversed order.
For each stage $s=1,2,\dots,\log_2 N$, with half-size $m=2^{s-1}$:
- for each block of length $2m$, and each $k=0,\dots,m-1$:
  - $W=W_{2m}^{k}=e^{-j2\pi k/(2m)}$
  - $t=W\cdot a[k+m]$
  - $a[k+m]=a[k]-t$
  - $a[k]=a[k]+t$
The array now holds $X[k]$ in natural order.

Each butterfly is one complex multiply and two adds; there are $(N/2)\log_2 N$ butterflies.

Q71. 8-point DFT of ${1,1,1,1,1,1,0,0}$

For $k\ne0$, sum the geometric series with $W_8=e^{-j\pi/4}$:

\[X[k]=\sum_{n=0}^{5}W_8^{nk}=\frac{1-W_8^{6k}}{1-W_8^{k}}=e^{-j5\pi k/8}\frac{\sin(3\pi k/4)}{\sin(\pi k/8)}\]

and $X[0]=6$. Evaluating:

$k$	$X[k]$
0	$6$
1	$-0.707-1.707j$
2	$1-j$
3	$0.707+0.293j$
4	$0$
5	$0.707-0.293j$
6	$1+j$
7	$-0.707+1.707j$

The symmetry $X[8-k]=X^{*}[k]$ for real input is visible across the table.

Q72. Define an impulse signal

The discrete unit impulse is $\delta[n]=1$ at $n=0$ and $0$ elsewhere. It has unit energy and acts as the identity of convolution. The continuous Dirac impulse $\delta(t)$ is a unit-area spike with $\int\delta(t)\,dt=1$ and the sifting property $\int x(t)\delta(t-t_0)\,dt=x(t_0)$. The impulse response $h[n]$ of a system is its response to $\delta[n]$ and completely characterizes an LTI system.

Q73. Compare FIR and IIR filters

FIR filters have a finite impulse response, no feedback, guaranteed stability, and can have exact linear phase. They need a high order for sharp transitions and therefore more computation and memory. IIR filters have an infinite impulse response, use feedback, need stability checks, and generally cannot achieve exact linear phase. They meet sharp specifications with a much lower order, so they are cheaper to compute.

Q74. Direct Form I for $y[n]=0.5y[n-1]-0.76y[n-2]+0.63y[n-3]+x[n]+0.875x[n-1]$

\[H(z)=\frac{1+0.875z^{-1}}{1-0.5z^{-1}+0.76z^{-2}-0.63z^{-3}}\]

Direct Form I has a forward chain and a feedback chain:

Forward taps: $b_0=1$, $b_1=0.875$.
Feedback taps: $a_1=-0.5$, $a_2=0.76$, $a_3=-0.63$ in the rearranged equation, i.e. the output is the sum of $x[n]$, $0.875x[n-1]$, $0.5y[n-1]$, $-0.76y[n-2]$, and $0.63y[n-3]$.
Six delay elements total (three for $x$, three for $y$); Direct Form II would use three.

Q75. What is signal processing? Analog versus digital signals

Signal processing is the deliberate modification or analysis of a signal to extract information, remove noise, enhance a feature, compress it, or prepare it for transmission.

Property	Analog signal	Digital signal
Time	continuous	discrete
Amplitude	continuous	quantized
Representation	voltage, current	binary numbers
Processing	analog circuits	processor and software
Noise	accumulates, drifts	can be regenerated exactly
Storage	tape, analog media	memory, disks
Flexibility	fixed by hardware	programmable

Academic, CSE-409 Digital Signal Processing

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