Let's solve both parts step by step. --- # Problem 6: Discrete-Time Fourier Transform (DTFT) The **DTFT** of a discrete signal \( x(n) \) is defined as: \[ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x(n) e^{-j\omega n} \] --- ## (a) \( x(n) = \left(\frac{1}{5}\right)^n u(n+3) \) ### **Step 1: Understand the Signal** - \( u(n+3) \) is the unit step function shifted left by 3, so it is 1 for \( n \geq -3 \). - So, \( x(n) = \left(\frac{1}{5}\right)^n \) for \( n \geq -3 \), and 0 otherwise. ### **Step 2: Express the Summation** \[ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x(n) e^{-j\omega n} = \sum_{n = -3}^{\infty} \left(\frac{1}{5}\right)^n e^{-j\omega n} \] ### **Step 3: Simplify the Summation** Factor terms: \[ X(e^{j\omega}) = \sum_{n = -3}^{\infty} \left(\frac{e^{-j\omega}}{5}\right)^n \] Let \( a = \frac{e^{-j\omega}}{5} \): \[ X(e^{j\omega}) = \sum_{n = -3}^{\infty} a^n \] ### **Step 4: Convert to a Standard Geometric Series** Break the sum: \[ \sum_{n = -3}^{\infty} a^n = \sum_{n=0}^{\infty} a^{n-3} = a^{-3} \sum_{n=0}^{\infty} a^n = a^{-3} \cdot \frac{1}{1-a}, \quad \text{for } |a| < 1 \] Recall \( a^{-3} = \left(\frac{e^{-j\omega}}{5}\right)^{-3} = 5^3 e^{j3\omega} = 125 e^{j3\omega} \). So, \[ X(e^{j\omega}) = 125 e^{j3\omega} \cdot \frac{1}{1 - \frac{e^{-j\omega}}{5}} \] \[ = \frac{125 e^{j3\omega}}{1 - \frac{e^{-j\omega}}{5}} \] Or, multiplying numerator and denominator by 5, \[ \boxed{ X(e^{j\omega}) = \frac{625 e^{j3\omega}}{5 - e^{-j\omega}} } \] --- ## (b) \( x(n) = \begin{cases} \left(\frac{1}{5}\right)^n & n = 0, 2, 4, \dots \\ 0 & \text{otherwise} \end{cases} \) ### **Step 1: Express the Summation** Nonzero only for even \( n \), so let \( n = 2k \), \( k = 0, 1, 2, \dots \): \[ X(e^{j\omega}) = \sum_{n = -\infty}^{\infty} x(n) e^{-j\omega n} = \sum_{k = 0}^{\infty} \left( \frac{1}{5} \right)^{2k} e^{-j\omega (2k)} \] \[ = \sum_{k=0}^{\infty} \left( \frac{1}{25} \right)^{k} e^{-j2\omega k} = \sum_{k=0}^{\infty} \left( \frac{e^{-j2\omega}}{25} \right)^{k} \] ### **Step 2: Sum the Geometric Series** Sum of geometric series: \[ \sum_{k=0}^{\infty} r^k = \frac{1}{1-r}, \quad |r| < 1 \] Here, \( r = \frac{e^{-j2\omega}}{25} \): \[ X(e^{j\omega}) = \frac{1}{1 - \frac{e^{-j2\omega}}{25}} \] ### **Final Boxed Answer:** \[ \boxed{ X(e^{j\omega}) = \frac{1}{1 - \frac{e^{-j2\omega}}{25}} } \] --- ## **Summary of Answers** ### **(a)** \[ \boxed{ X(e^{j\omega}) = \frac{625 e^{j3\omega}}{5 - e^{-j\omega}} } \] ### **(b)** \[ \boxed{ X(e^{j\omega}) = \frac{1}{1 - \frac{e^{-j2\omega}}{25}} } \]