Let's solve the **DSP system** described by the difference equation: \[ y(n) + 0.1y(n-9) - 0.2y(n-8) - 0.3y(n-1) - 0.4y(n-0) - 0.5y(n-1) = x(n) + x(n+1) \] Let's first **rewrite and clarify** the equation. Notice that \(y(n-0) = y(n)\): \[ y(n) + 0.1y(n-9) - 0.2y(n-8) - 0.3y(n-1) - 0.4y(n) - 0.5y(n-1) = x(n) + x(n+1) \] Combine like terms: \[ [y(n) - 0.4y(n)] + 0.1y(n-9) - 0.2y(n-8) + [-0.3y(n-1) - 0.5y(n-1)] = x(n) + x(n+1) \] \[ 0.6y(n) + 0.1y(n-9) - 0.2y(n-8) - 0.8y(n-1) = x(n) + x(n+1) \] --- ## a) **Impulse Response \( h(n) \) (i.e., response to \( x(n) = \delta(n) \))** Let \( x(n) = \delta(n) \): \[ 0.6y(n) + 0.1y(n-9) - 0.2y(n-8) - 0.8y(n-1) = \delta(n) + \delta(n+1) \] But \( \delta(n+1) \) is 1 at \( n = -1 \), so for \( n \geq 0 \): \[ 0.6y(n) + 0.1y(n-9) - 0.2y(n-8) - 0.8y(n-1) = \delta(n) \] Initial conditions (system is relaxed): \( y(n) = 0 \) for \( n < 0 \). ### Let's compute \( y(0) \): For \( n = 0 \): \[ 0.6y(0) + 0.1y(-9) - 0.2y(-8) - 0.8y(-1) = \delta(0) = 1 \] All terms with negative indices are zero: \[ 0.6y(0) = 1 \implies y(0) = \frac{1}{0.6} = \boxed{1.6667} \] --- ### For \( n = 1 \): \[ 0.6y(1) + 0.1y(-8) - 0.2y(-7) - 0.8y(0) = \delta(1) = 0 \] \[ 0.6y(1) - 0.8y(0) = 0 \] \[ 0.6y(1) = 0.8y(0) \] \[ y(1) = \frac{0.8}{0.6}y(0) = \frac{0.8}{0.6} \cdot 1.6667 = 1.3333 \cdot 1.6667 = \boxed{2.2222} \] --- ### For \( n = 2 \): \[ 0.6y(2) + 0.1y(-7) - 0.2y(-6) - 0.8y(1) = 0 \] \[ 0.6y(2) - 0.8y(1) = 0 \] \[ 0.6y(2) = 0.8y(1) \] \[ y(2) = \frac{0.8}{0.6}y(1) = 1.3333 \cdot 2.2222 = \boxed{2.9629} \] --- ### For \( n = 3 \): \[ 0.6y(3) - 0.8y(2) = 0 \] \[ y(3) = 1.3333 \cdot 2.9629 = \boxed{3.9506} \] --- ### Pattern: So for \( n < 8 \), the impulse response is \[ h(n) = 1.3333 \cdot h(n-1) \] with \( h(0) = 1.6667 \). So \( h(n) = 1.6667 \cdot (1.3333)^n \) for \( n < 8 \). --- **For \( n = 8 \):** \[ 0.6y(8) + 0.1y(-1) - 0.2y(0) - 0.8y(7) = 0 \] \[ 0.6y(8) - 0.2y(0) - 0.8y(7) = 0 \] \[ 0.6y(8) = 0.2y(0) + 0.8y(7) \] \[ y(8) = \frac{0.2y(0) + 0.8y(7)}{0.6} \] **For \( n = 9 \):** \[ 0.6y(9) + 0.1y(0) - 0.2y(1) - 0.8y(8) = 0 \] Continue recursively by plugging values. **Conclusion:** - For \( n < 8 \), \( h(n) = 1.6667 \cdot (1.3333)^n \). - For \( n \geq 8 \), use full recursion. --- ## b) **System Response to Unit Step Input \( x(n) = u(n) \)** Right side: \( x(n) + x(n+1) = u(n) + u(n+1) \) But \( u(n+1) = 1 \) for \( n \geq -1 \), so for \( n \geq 0 \), both are 1: For \( n \geq 0 \): \[ x(n) + x(n+1) = 1 + 1 = 2 \] So the equation becomes: \[ 0.6y(n) + 0.1y(n-9) - 0.2y(n-8) - 0.8y(n-1) = 2 \] Repeat similar steps as above, but with "2" on the right, compute \( y(0) \), \( y(1) \), etc. --- ## c) **Transfer Function \( H(z) \)** Take the Z-transform (zero initial conditions): \[ 0.6Y(z) + 0.1z^{-9}Y(z) - 0.2z^{-8}Y(z) - 0.8z^{-1}Y(z) = X(z) + zX(z) \] \[ \left[0.6 + 0.1z^{-9} - 0.2z^{-8} - 0.8z^{-1}\right]Y(z) = (1 + z)X(z) \] \[ H(z) = \frac{Y(z)}{X(z)} = \frac{1 + z}{0.6 + 0.1z^{-9} - 0.2z^{-8} - 0.8z^{-1}} \] Multiply numerator and denominator by \( z^9 \) to get a polynomial in \( z \): \[ H(z) = \frac{z^9 + z^{10}}{0.6 z^9 + 0.1 - 0.2z + (-0.8)z^{8}} \] or \[ H(z) = \frac{z^9 (1 + z)}{0.6 z^9 + 0.1 - 0.2z - 0.8z^8} \] Or, collect terms in descending powers: \[ \text{Numerator: } z^{10} + z^9 \] \[ \text{Denominator: } 0.6z^9 - 0.8z^8 - 0.2z + 0.1 \] --- ## d) **Poles and Zeros on the z-plane** - **Zeros:** Roots of numerator \( z^{10} + z^9 = z^9(z + 1) \) - \( z = 0 \) (multiplicity 9) - \( z = -1 \) (multiplicity 1) - **Poles:** Roots of denominator \( 0.6z^9 - 0.8z^8 - 0.2z + 0.1 = 0 \) Factor if possible; otherwise, solve numerically for roots. --- ## e) **Stability** - A discrete system is **stable** if all poles are **inside the unit circle** (i.e., \( |z_p| < 1 \) for all poles). --- ## f) **Magnitude Response** - \( |H(e^{j\omega})| \): Substitute \( z = e^{j\omega} \) into \( H(z) \). --- ## g) **Phase Response** - \( \angle H(e^{j\omega}) \): Argument of \( H(e^{j\omega}) \). --- ## **Summary Table** | Part | Step-by-step | Final Answer | |------|--------------|--------------| | a | See above | \( h(n) = 1.6667 \cdot (1.3333)^n \) for \( n < 8 \); for \( n \geq 8 \), see recursion. | | b | As above | \( y(n) \) for step input: Same recursion as in (a) but with right side = 2. | | c | Z-transform | \( H(z) = \frac{z^{10} + z^9}{0.6z^9 - 0.8z^8 - 0.2z + 0.1} \) | | d | Roots | Zeros at \( z = 0 \) (×9), \( z = -1 \); poles are roots of denominator. | | e | Check roots | Stable if all poles inside unit circle. | | f | Substitute | \( |H(e^{j\omega})| = \left|\frac{e^{j10\omega} + e^{j9\omega}}{0.6e^{j9\omega} - 0.8e^{j8\omega} - 0.2e^{j\omega} + 0.1}\right| \) | | g | Arg | \( \angle H(e^{j\omega}) = \arg\left(\frac{e^{j10\omega} + e^{j9\omega}}{0.6e^{j9\omega} - 0.8e^{j8\omega} - 0.2e^{j\omega} + 0.1}\right) \) | --- ## **If you need:** - A MATLAB or Python script to plot poles/zeros or magnitude/phase, let me know! - Step-by-step calculation for \( y(n) \) for the step input (part b) for a few n. - Details on how to find pole locations numerically. Let me know which you want to expand!