THEORY EXAMINATION (SEM–IV) 2016-17 SIGNALS AND SYSTEMS

Course: B.Tech (EE / EC / ECE / EN)
Subject Code: EEC404
Subject Title: Signals and Systems
Exam Type: Theory
Duration: 3 Hours
Maximum Marks: 100

SECTION – A (10 × 2 = 20 Marks)

Short theoretical questions on key definitions and properties

(c) Causality & Stability

Causal System: Output depends only on present and past inputs.
Example: y(t)=x(t−1)y(t) = x(t-1)y(t)=x(t−1).

Stable System: BIBO stable if bounded input gives bounded output.
Condition: ∑∣h[n]∣<∞\sum |h[n]| < \infty∑∣h[n]∣<∞.

(d) Discrete-Time Signal Sketch

Given:

x(n)=2−n,−3<n<3x(n) = 2^{-n}, \quad -3 < n < 3x(n)=2−n,−3<n<3

and Y(n)=x(n)+u(−n+2)Y(n) = x(n) + u(-n+2)Y(n)=x(n)+u(−n+2).
Requires plotting of exponential decay and step function for specific nnn range.

(e) Relation Between Laplace and Fourier Transforms

Fourier Transform is a special case of Laplace Transform when the region of convergence (ROC) includes the imaginary axis s=jωs = jωs=jω:

X(jω)=X(s)∣s=jωX(jω) = X(s)\big|_{s=jω}X(jω)=X(s)​s=jω​ 

(f) Parseval’s Theorem

Energy of a signal is same in time and frequency domain:

∫∣x(t)∣2dt=∫∣X(f)∣2df\int |x(t)|^2 dt = \int |X(f)|^2 df∫∣x(t)∣2dt=∫∣X(f)∣2df

Used for power and energy calculations.

(g) Solving Difference Equation (Z-transform method)

x(n−2)−9x(n−1)+18x(n)=0x(n-2) - 9x(n-1) + 18x(n) = 0x(n−2)−9x(n−1)+18x(n)=0

Given: x(−1)=1,x(−2)=9x(-1)=1, x(-2)=9x(−1)=1,x(−2)=9.
Apply Z-transform, solve algebraically, and find inverse using partial fraction expansion.

(h) Convolution Integral

For continuous-time systems:

y(t)=x(t)∗h(t)=∫−∞∞x(τ)h(t−τ)dτy(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(τ)h(t-τ)dτy(t)=x(t)∗h(t)=∫−∞∞​x(τ)h(t−τ)dτ

Represents output response of an LTI system.

SECTION – C (2 × 15 = 30 Marks)

Comprehensive and application-level questions

Q3. Energy Spectral Density (ESD)

Definition: Measures energy distribution over frequency.

• Sx(f)=∣X(f)∣2S_x(f) = |X(f)|^2Sx​(f)=∣X(f)∣2

For x(t)=e−tu(t)x(t) = e^{-t}u(t)x(t)=e−tu(t):

• X(f)=11+j2πf⇒Sx(f)=11+(2πf)2X(f) = \frac{1}{1 + j2πf} \Rightarrow S_x(f) = \frac{1}{1 + (2πf)^2}X(f)=1+j2πf1​⇒Sx​(f)=1+(2πf)21​

Auto-correlation Function:

• Rx(τ)=∫x(t)x(t+τ)dt=12e−∣τ∣R_x(τ) = \int x(t)x(t+τ)dt = \frac{1}{2}e^{-|τ|}Rx​(τ)=∫x(t)x(t+τ)dt=21​e−∣τ∣

Q4. One-Sided Laplace Transform

Definition:

• X(s)=∫0∞x(t)e−stdtX(s) = \int_0^{\infty} x(t)e^{-st}dtX(s)=∫0∞​x(t)e−stdt

for t≥0t ≥ 0t≥0 (causal systems).

Example:

• x(t)=e−3tu(t)+e−2tu(t)x(t) = e^{-3t}u(t) + e^{-2t}u(t)x(t)=e−3tu(t)+e−2tu(t) X(s)=1s+3+1s+2X(s) = \frac{1}{s+3} + \frac{1}{s+2}X(s)=s+31​+s+21​

ROC: Re(s)>−2Re(s) > -2Re(s)>−2.

Q5. DTFT and CT Low-Pass Filter

(a) Discrete-Time Fourier Transform (DTFT):

• X(ejω)=∑n=−∞∞x[n]e−jωnX(e^{jω}) = \sum_{n=-∞}^{∞} x[n]e^{-jωn}X(ejω)=n=−∞∑∞​x[n]e−jωn

Converts discrete signals into continuous frequency representation.

(b) First-Order Continuous-Time (CT) Low-Pass Filter:

• H(s)=11+sRCH(s) = \frac{1}{1 + sRC}H(s)=1+sRC1​

Cutoff Frequency: ωc=1RCω_c = \frac{1}{RC}ωc​=RC1​

Frequency Response: Allows low frequencies, attenuates high frequencies.

 Summary

This Signals and Systems (EEC404) paper evaluates: