Question: Department of Electrical and Electronic Engineering 4th Year 1st Semester Final Examination - 2024 Course Code: EEE 4131 Course Title: Digital Signal Processing Time: 3 hours Full Marks: 70 (N. B: Answer any five of the following questions. Figures in the right margin indicate marks allocated for each question. You should answer all subquestions of the same question consecutively.) 1. a) How is "Digital Signal Processing" defined? Explain the fundamental difference between analog signals and digital signals. 05 b) Consider the analog signal: x_a(t) = 3 cos 2000\pi t + 5 sin 6000\pi t + 10 cos 14000\pi t i. What is the Nyquist rate for this signal? ii. Assume now that we sample this signal using a sampling rate F_s = 9000 samples/s. What is the discrete-time signal obtained after sampling? 2+4 c) Explain what aliasing is and describe how it can be avoided in a digital signal processing system. 03 2. a) A triangular pulse signal x(t) is depicted in the figure. Sketch each of the following signals derived from x(t): i. x(5t - 2) ii. x(3(t + 6)) iii. x(2(t - 1)) 06 b) Describe the basic operation on dependent and independent signal with proper notation and explanation. 03 c) For the given transfer function. H(z) = (1 + 0.25 z^{-1}) / (1 - 0.1 z^{-1} + 3 z^{-2}) i. Plot the pole-zero diagram of the H(z). ii. Find a stable impulse response h(n). 05 3. a) Draw the direct form - I structure of the following system: y(n) = - \sum_{k=1}^{5} a_k y(n-k) + \sum_{k=0}^{3} b_k x(n-k) 08 b) Compute the filter output of below configuration with DTFT method. x[n] = h[n] = { 2 for 0 \leq n \leq 3 { 0 Otherwise 06 4. a) Explain the following terms with its application in DSP: i) Convolution ii) Auto-Correlation iii) Modulation 04 b) To understand the process, let us study the analog system shown in the following figure. It is desired to find the output when the input is a unit step function v(t) = u(t), and the system is initially relaxed at t = 0 or equivalently the system has zero initial conditions. Graph/figure: top-right; a right-margin point label shows "05" near a circled mark. Graph/figure: center-left; a small sketch of x(t) shows a triangular/ramp-like signal starting at (t=0, x=0), rising linearly to x=2 at t=2, then dropping to x=1 and staying constant until t=4, ending there. Graph/figure: bottom-left; an analog circuit with a source u(t), an inductor labeled L = 1 H in the top branch, and a resistor labeled R = 0.5 \Omega on the right branch, with output v_o(t) taken across the resistor/output node.User: give the solution

1. **Problem Statement:** Define Digital Signal Processing (DSP), distinguish analog and digital signals, analyze a given analog signal for Nyquist rate and sampled discrete-time signal, and explain aliasing. 2. **Definition and Difference:** - Digital Signal Processing is the manipulation of signals after converting them into digital form. - Analog signals are continuous in time and amplitude, while digital signals are discrete in time and amplitude. 3. **Given Signal:** $$x_a(t) = 3 \cos(2000\pi t) + 5 \sin(6000\pi t) + 10 \cos(14000\pi t)$$ 4. **Nyquist Rate:** - Nyquist rate is twice the maximum frequency component. - Frequencies are: - $f_1 = \frac{2000\pi}{2\pi} = 1000$ Hz - $f_2 = \frac{6000\pi}{2\pi} = 3000$ Hz - $f_3 = \frac{14000\pi}{2\pi} = 7000$ Hz - Maximum frequency $f_{max} = 7000$ Hz - Nyquist rate $= 2 \times 7000 = 14000$ samples/s 5. **Sampling at $F_s = 9000$ samples/s:** - Sampling frequency $F_s = 9000$ Hz < Nyquist rate $14000$ Hz, so aliasing occurs. - Discrete-time frequency $\omega = 2\pi \frac{f}{F_s}$ - Calculate discrete frequencies: - $\omega_1 = 2\pi \frac{1000}{9000} = \frac{2\pi}{9}$ - $\omega_2 = 2\pi \frac{3000}{9000} = \frac{2\pi}{3}$ - $\omega_3 = 2\pi \frac{7000}{9000} = \frac{14\pi}{9}$ - Since $\omega_3 > \pi$, aliasing occurs: - Aliased frequency $\omega_3' = 2\pi - \frac{14\pi}{9} = \frac{4\pi}{9}$ - Discrete-time signal: $$x[n] = 3 \cos\left(\frac{2\pi}{9} n\right) + 5 \sin\left(\frac{2\pi}{3} n\right) + 10 \cos\left(\frac{4\pi}{9} n\right)$$ 6. **Aliasing Explanation:** - Aliasing is the distortion that occurs when sampling frequency is less than twice the maximum frequency. - It causes higher frequencies to appear as lower frequencies. - Avoided by sampling at or above Nyquist rate and using anti-aliasing filters. --- 7. **Problem Statement:** Given triangular pulse $x(t)$, sketch $x(5t-2)$, $x(3(t+6))$, $x(2(t-1))$. 8. **Signal Transformations:** - $x(5t-2)$: compress time by 5, shift right by $\frac{2}{5}$. - $x(3(t+6))$: compress time by 3, shift left by 6. - $x(2(t-1))$: compress time by 2, shift right by 1. (Sketches would show these time scalings and shifts.) --- 9. **Problem Statement:** Describe basic operations on dependent and independent signals. 10. **Explanation:** - Dependent variable changes with independent variable. - Independent variable is input or time. - Notation: $y(t) = f(x(t))$ where $x(t)$ is independent, $y(t)$ dependent. --- 11. **Problem Statement:** Given transfer function $$H(z) = \frac{1 + 0.25 z^{-1}}{1 - 0.1 z^{-1} + 3 z^{-2}}$$ - Plot pole-zero diagram. - Find stable impulse response $h(n)$. 12. **Pole-Zero Calculation:** - Zeros: roots of numerator $1 + 0.25 z^{-1} = 0 \Rightarrow z = -0.25$ - Poles: roots of denominator $1 - 0.1 z^{-1} + 3 z^{-2} = 0$ Multiply by $z^2$: $$z^2 - 0.1 z + 3 = 0$$ - Solve quadratic: $$z = \frac{0.1 \pm \sqrt{0.01 - 12}}{2} = \frac{0.1 \pm j\sqrt{11.99}}{2}$$ - Poles are complex conjugates with magnitude $>1$, system unstable. 13. **Impulse Response:** - Since poles outside unit circle, system is unstable. - Stable impulse response does not exist unless poles inside unit circle. --- 14. **Problem Statement:** Draw direct form-I structure for $$y(n) = - \sum_{k=1}^5 a_k y(n-k) + \sum_{k=0}^3 b_k x(n-k)$$ 15. **Explanation:** - Direct form-I has two sections: feedforward (b coefficients) and feedback (a coefficients). - Feedback uses past outputs, feedforward uses current and past inputs. --- 16. **Problem Statement:** Compute filter output with DTFT method for $$x[n] = h[n] = \begin{cases} 2 & 0 \leq n \leq 3 \\ 0 & \text{otherwise} \end{cases}$$ 17. **DTFT Calculation:** - DTFT of $x[n]$: $$X(\omega) = \sum_{n=0}^3 2 e^{-j \omega n} = 2 \sum_{n=0}^3 e^{-j \omega n} = 2 \frac{1 - e^{-j4\omega}}{1 - e^{-j\omega}}$$ - Since $y[n] = x[n] * h[n]$ and $x[n] = h[n]$, output is convolution of $x[n]$ with itself. --- 18. **Problem Statement:** Explain convolution, auto-correlation, modulation with DSP applications. 19. **Definitions:** - Convolution: combines two signals to produce output; used in filtering. - Auto-correlation: measures similarity of signal with delayed version; used in signal detection. - Modulation: varies signal properties to transmit data; used in communication. --- 20. **Problem Statement:** Find output $v_o(t)$ of analog circuit with input $v(t) = u(t)$, $L=1$ H, $R=0.5 \Omega$, zero initial conditions. 21. **Circuit Analysis:** - Differential equation: $$L \frac{di}{dt} + R i = v(t)$$ - With $v(t) = u(t)$ and $L=1$, $R=0.5$: $$\frac{di}{dt} + 0.5 i = u(t)$$ 22. **Solve ODE:** - Homogeneous solution: $$i_h(t) = K e^{-0.5 t}$$ - Particular solution for $t \geq 0$: $$i_p = \text{constant}$$ - Substitute into ODE: $$0 + 0.5 i_p = 1 \Rightarrow i_p = 2$$ - General solution: $$i(t) = 2 + K e^{-0.5 t}$$ - Initial condition $i(0) = 0$: $$0 = 2 + K \Rightarrow K = -2$$ - Final current: $$i(t) = 2 (1 - e^{-0.5 t}) u(t)$$ 23. **Output voltage across resistor:** $$v_o(t) = R i(t) = 0.5 \times 2 (1 - e^{-0.5 t}) u(t) = (1 - e^{-0.5 t}) u(t)$$ **Final answer:** $$v_o(t) = (1 - e^{-0.5 t}) u(t)$$