Compiled DSP Questions: Answers 1–25 (Signals, Sampling, Quantization, Statistics)

Worked answers to questions 1-25 of the compiled CSE-409 DSP question list, with links to the full theory posts.

Posted Mar 18, 2026

By Sadman Ishtiak

views 9 min read

How This Post Fits the Series

The question list this post answers is the Compiled DSP Questions for CSE-409 page. The numbering below matches that page exactly. Every answer is short enough to revise from, and the “Read more” links point to the post that derives the same idea in full. Numerical answers from the past exams are also worked in CSE 409 Past Exams and CSE 410 Lab: Worked Solutions.

Q1. Define and classify signals with examples

A signal is a function that carries information about the state of a physical system, usually as amplitude varying with time or space. A signal is classified by:

Time axis: continuous-time $x(t)$ (defined for every $t$) versus discrete-time $x[n]$ (defined only at integer $n$).
Amplitude axis: analog (continuous values) versus digital (quantized values).
Predictability: deterministic (exact formula) versus random (statistical description).
Repetition: periodic versus aperiodic.
Energy: energy signal ($0<E<\infty$) versus power signal ($0<P<\infty$).
Symmetry: even, odd, or neither.
Causality: causal ($x[n]=0$ for $n<0$) versus non-causal.

Examples: a microphone voltage is analog continuous-time; a sampled audio track is digital discrete-time; a sinusoid is deterministic and periodic; thermal noise is random.

Read more: What Is a Signal? and Signal Classifications.

Q2. Continuous-time versus discrete-time signals

Property	Continuous-time $x(t)$	Discrete-time $x[n]$
Independent variable	real $t$	integer $n$
Defined	every instant	only at sample instants
Amplitude	continuous	can be continuous or quantized
Notation	parentheses	square brackets
Typical source	sensor output	sampler / computer

Diagram: a smooth curve through all points versus stems only at $n=0,\pm1,\pm2,\dots$ A discrete-time signal can come from sampling, $x[n]=x_a(nT_s)$, or be defined directly by a formula such as $x[n]=a^n u[n]$.

Q3. Sketch $x(t)=\sin(\pi t)$ for $-5\le t\le 5$

$\sin(\pi t)$ has angular frequency $\pi$ rad/s, so $F=1/2$ Hz and period $T=2$ s. It is zero at every integer $t$, reaches $+1$ at $t=0.5,2.5$ and $-1$ at $t=-0.5,-3.5$. Between $t=-5$ and $t=5$ there are five full periods. The sketch is a sine wave crossing the axis at $-5,-4,\dots,5$ and peaking halfway between crossings.

Q4. Three major real-world applications of DSP

Audio and speech: noise reduction, echo cancellation, speech recognition, MP3/AAC compression.
Medical imaging: CT/MRI reconstruction, ultrasound beamforming, image denoising and enhancement.
Radar, sonar and wireless communication: matched filtering, pulse compression, modulation and channel equalization.

Q5. DSP in speech recognition

The microphone signal is pre-emphasized, split into short frames (20–30 ms), windowed, and transformed with the FFT. The spectrum is converted to mel-scale filter-bank energies and then to MFCCs (or fed directly to a neural network). Voice activity detection, feature normalization and decoding follow. DSP supplies the front end; the acoustic model maps features to phonemes or words.

Q6. DSP in medical image enhancement

DSP removes noise (median and Gaussian filtering), improves contrast (histogram equalization, CLAHE), sharpens edges (unsharp masking, high-pass filtering), interpolates to a finer grid, and compresses for storage. In CT and MRI, the reconstruction itself is an inverse Fourier / filtered back-projection operation, which is pure DSP.

Q7. Why digital filters are preferred over analog in communication

Programmable: change the response by changing coefficients, no hardware redesign.
Reproducible: no component tolerance, drift, or ageing.
Exact linear phase is possible (FIR), so no phase distortion.
Arbitrary specifications and adaptive filtering are practical.
Coefficients can be stored, versioned and copied exactly.
Lower cost for complex, high-order responses.

Q8. $x_a(t)=3\sin 300\pi t$

The angular frequency is $300\pi$, so $F=150$ Hz.

a) Minimum sampling rate. $F_s^{\min}=2F=300$ Hz.

b) $F_s=250$ Hz. Substitute $t=nT_s=n/250$:

\[x[n]=3\sin\left(2\pi\frac{150}{250}n\right)=3\sin\left(\frac{6\pi}{5}n\right)\]

Since $150>F_s/2=125$ Hz, this is an aliased tone. Using $\sin(6\pi n/5)=\sin(6\pi n/5-2\pi n)=\sin(-4\pi n/5)$:

\[x[n]=-3\sin\left(\frac{4\pi}{5}n\right)\]

which behaves like a $100$ Hz tone after folding, because $250-150=100$ Hz.

Q9. Symmetric and anti-symmetric signals

A signal is even (symmetric) if $x[-n]=x[n]$ for all $n$: examples $\delta[n]$, $\cos\omega n$, $u[n]+u[-n]$ (apart from the origin convention).

A signal is odd (anti-symmetric) if $x[-n]=-x[n]$: examples $\sin\omega n$, $n$, and any signal with $x[0]=0$ built from odd parts.

Every signal decomposes as $x[n]=x_e[n]+x_o[n]$ with

\[x_e[n]=\frac{x[n]+x[-n]}{2},\qquad x_o[n]=\frac{x[n]-x[-n]}{2}\]

Q10. Aliasing in the frequency domain

Sampling multiplies the spectrum by an impulse train, producing copies of $X_a(F)$ centred at $kF_s$. If $F_s<2F_{\max}$, neighbouring copies overlap. A component above $F_s/2$ then appears at $F_s-F$, an image at a lower frequency that cannot be separated from a genuine low-frequency component. Larger overlap means more irreversible distortion.

Q11. Quantization and its significance

Quantization maps a continuous-amplitude sample to one of $L=2^B$ discrete levels. It is significant because digital hardware can only store and process finite-precision numbers; quantization makes the signal digital, at the cost of a quantization error that behaves like additive noise.

Q12. Quantization error of a uniform quantizer

For a uniform quantizer with step

\[\Delta=\frac{x_{\max}-x_{\min}}{L},\qquad L=2^B\]

the output is $\hat{x}=Q(x)$. The error is $e=\hat{x}-x$. With rounding, $e$ lies in $[-\Delta/2,\Delta/2]$. Assuming $e$ is uniform in that interval,

\[E[e]=\frac{1}{\Delta}\int_{-\Delta/2}^{\Delta/2}e\,de=0\] \[E[e^2]=\frac{1}{\Delta}\int_{-\Delta/2}^{\Delta/2}e^2\,de=\frac{\Delta^2}{12}\]

So the quantization noise power is $\sigma_e^2=\Delta^2/12$. For a full-scale sinusoid the signal power is $A^2/2$ with $A=x_{\max}$, giving $\text{SQNR}\approx 6.02B+1.76$ dB.

Q13. Step size and quantized value for a 16-level quantizer

Range $-10$ V to $+10$ V, so $x_{\max}-x_{\min}=20$ V and $L=16$:

\[\Delta=\frac{20}{16}=1.25\ \text{V}\]

For $x=3.8$ V, the code index measured from $-10$ V is

\[k=\frac{3.8-(-10)}{1.25}=11.04\approx 11\] \[\hat{x}=-10+11(1.25)=3.75\ \text{V}\]

The error is $3.75-3.8=-0.05$ V, well inside the $\pm0.625$ V bound.

Q14. Time scaling and its effects

Time scaling replaces $t$ by $at$: $x(at)$.

$a>1$ compresses the signal in time (it plays faster), so its spectrum expands in frequency.
$0<a<1$ expands the signal in time, so its spectrum compresses.
For discrete-time signals only rational factors are meaningful: $x[2n]$ is decimation (keep every second sample), $x[n/2]$ is interpolation (insert zeros then filter).

Q15. Signal shifting

Replacing $t$ by $t-t_0$ with $t_0>0$ shifts the waveform right (delay); replacing $t$ by $t+t_0$ shifts it left (advance). In discrete time, $x[n-k]$ is delayed by $k$ samples and $x[n+k]$ is advanced by $k$. A shift changes only the time origin, never the shape or the amplitude.

Q16. For $x[n]={2,1,\mathbf{0},-1,-2}$, plot $x[2n]$ and $x[n/2]$

The bold $0$ is at $n=0$, so the sequence occupies $n=-2,-1,0,1,2$.

$x[2n]$ keeps only even-indexed samples:

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

All other values are zero. This is down-sampling by 2.

$x[n/2]$ is nonzero only when $n/2$ is an integer, i.e. $n$ even:

$n$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$x[n/2]$	$2$	$0$	$1$	$0$	$0$	$0$	$-1$	$0$	$-2$

This is up-sampling by 2 with zero insertion.

Q17. Difference between DFT and DTFT

Property	DTFT	DFT
Frequency variable	continuous $\omega$	discrete $k$
Length of input	any, often infinite	finite $N$
Output	continuous function, periodic with $2\pi$	$N$ samples, periodic with $N$
Computation	integral/sum over all $n$	finite sum, computable exactly
Relation	reference transform	samples DTFT at $\omega=2\pi k/N$

The DFT is the computable, finite version of the DTFT: $X[k]=X(e^{j\omega})$ at $\omega=2\pi k/N$.

Q18. Compute the 4-point DFT of ${0,1,2,3}$

With $W_4=e^{-j2\pi/4}=-j$,

\[X[k]=\sum_{n=0}^{3}x[n]W_4^{nk}\]

$X[0]=0+1+2+3=6$
$X[1]=0+1(-j)+2(-1)+3(j)=-2+2j$
$X[2]=0+1(-1)+2(1)+3(-1)=-2$
$X[3]=0+1(j)+2(-1)+3(-j)=-2-2j$

So $X={6,\,-2+2j,\,-2,\,-2-2j}$.

Q19. Advantages of FFT over DFT

The DFT costs $N^2$ complex multiplications; the radix-2 FFT costs $(N/2)\log_2 N$. For $N=64$ this is $4096$ versus $192$, a factor of $21.33$. Fewer operations mean faster execution, real-time feasibility, lower power, and less rounding noise from repeated arithmetic. The result is mathematically identical to the DFT.

Q20. Mean and median of ${1,3,5,7,9}$

\[\mu=\frac{1+3+5+7+9}{5}=5\]

Sorted order is already $1,3,5,7,9$, so the median is the third value, $5$. The set is symmetric, hence mean equals median.

Q21. Variance and standard deviation for ${2,4,6,8}$

\[\mu=\frac{2+4+6+8}{4}=5\] \[\sigma^2=\frac{1}{N}\sum (x_i-\mu)^2=\frac{9+1+1+9}{4}=5\] \[\sigma=\sqrt{5}\approx2.236\]

If the four values are treated as a sample, use $N-1$:

\[s^2=\frac{20}{3}\approx6.667,\qquad s\approx2.582\]

Q22. Skewness and kurtosis for ${1,2,3,4,9}$

Mean is $3.8$. Deviations are $-2.8,-1.8,-0.8,0.2,5.2$; sum of squares is $38.8$, so the population variance is $7.76$ and $\sigma\approx2.786$.

Standardized third moment:

\[\text{skewness}=\frac{1}{N}\sum\left(\frac{x_i-\mu}{\sigma}\right)^3\approx1.04\]

Positive skew: the large value $9$ pulls the right tail out. The median, $3$, is below the mean, $3.8$, which is the usual signature of right skew.

Standardized fourth moment minus 3:

\[\text{excess kurtosis}=\frac{1}{N}\sum\left(\frac{x_i-\mu}{\sigma}\right)^4-3\approx-0.33\]

That is close to a normal distribution (mesokurtic); the single outlier does not make the peak heavy enough for a positive excess.

Q23. Analog versus digital filters

Property	Analog filter	Digital filter
Hardware	resistors, capacitors, op-amps	processor plus program
Response change	redesign components	change coefficients
Drift and tolerance	yes	no
Precision	limited by components	set by word length
Linear phase	hard	straightforward with FIR
Frequency range	can reach RF	limited by sampling rate
Adaptive	difficult	natural

Q24. IIR design by the bilinear transform

State the digital specifications: passband edge $\omega_p$, stopband edge $\omega_s$, and ripples.
Pre-warp each edge with $\Omega=(2/T)\tan(\omega/2)$, which compensates the frequency compression of the mapping.
Design an analog prototype $H_a(s)$ (Butterworth, Chebyshev, elliptic) to the pre-warped specifications.
Substitute

\[s=\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}\]

to obtain $H(z)$.

Verify the digital response. The left half $s$-plane maps inside the unit circle, so a stable analog filter yields a stable digital filter.

Q25. FIR versus IIR filters

Property	FIR	IIR
Impulse response	finite	infinite
Feedback	none	yes
Stability	always stable	must check poles
Linear phase	easy (symmetric coefficients)	generally not exact
Order for a sharp response	high	low
Computation and memory	more	less
Design	windows, frequency sampling, Parks-McClellan	analog prototype plus mapping