Impulse response of recursive DT system

In summary, the impulse response h[n] for the given equation can be written as h[n] = (10+5u[n-1])(0.9)^n, where u[n] is the unit step function. This expression can be obtained by using the z-transform and partial fraction decomposition.
  • #1
dionysian
53
1

Homework Statement



[tex]y[n] - 1.8\cos (\frac{\pi }{{16}})y[n - 1] + 0.81y[n - 2] = x[n] + 0.5x[n - 1][/tex]
Determine the impulse response [tex]h[n][/tex] by calculating the zero-state response with [tex]x[n] = \delta [n][/tex]

Homework Equations



[tex]y[n] - 1.8\cos (\frac{\pi }{{16}})y[n - 1] + 0.81y[n - 2] = x[n] + 0.5x[n - 1][/tex]

The Attempt at a Solution



[tex]y[n] - 1.8\cos (\frac{\pi }{{16}})y[n - 1] + 0.81y[n - 2] = x[n] + 0.5x[n - 1][/tex]
[tex]y[n] = 1.8\cos (\frac{\pi }{{16}})y[n - 1] - 0.81y[n - 2] + x[n] + 0.5x[n - 1][/tex]
[tex]h[n] = 1.8\cos (\frac{\pi }{{16}})y[n - 1] - 0.81y[n - 2] + \delta [n] + 0.5\delta [n - 1][/tex]
The zero-state response means
[tex]\begin{array}{l}
y[- 1] = 0 \\
y[- 2] = 0 \\
\end{array}[/tex]
So my question is how can I write out the impulse response explicitly in one expression. I think I could calculate it by [tex]h[0],h[1],h[2]...[/tex] but I want to write it out as just one expression but the recursive terms[tex]y[n - 1][/tex] and [tex]y[n - 2][/tex] are kind of throwing me off.
 
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  • #2


To write the impulse response explicitly in one expression, you can use the z-transform. The z-transform of the given equation is:

Y(z) - 1.8cos(\frac{\pi}{16})z^{-1}Y(z) + 0.81z^{-2}Y(z) = X(z) + 0.5z^{-1}X(z)

Using the property of the z-transform that relates the output to the input for a linear time-invariant system, we get:

H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1-1.8cos(\frac{\pi}{16})z^{-1} + 0.81z^{-2}} + \frac{0.5z^{-1}}{1-1.8cos(\frac{\pi}{16})z^{-1} + 0.81z^{-2}}

Using partial fraction decomposition, we can write the above expression as:

H(z) = \frac{A}{1-\alpha z^{-1}} + \frac{B}{1-\beta z^{-1}}

where A and B are constants and \alpha and \beta are the roots of the denominator polynomial. In this case, \alpha = 0.9 and \beta = 0.9. Solving for A and B, we get:

A = \frac{1}{1-0.9} = 10
B = \frac{0.5}{1-0.9} = 5

Therefore, the impulse response in one expression is:

h[n] = 10(0.9)^n + 5(0.9)^n u[n-1]
 

FAQ: Impulse response of recursive DT system

1. What is the impulse response of a recursive DT system?

The impulse response of a recursive DT system is the output of the system when an impulse input is applied, which is a single sample with a value of 1 followed by zeros. It represents how the system will respond to any input signal.

2. How is the impulse response of a recursive DT system calculated?

The impulse response of a recursive DT system is calculated by applying an impulse input to the system and observing the output. This can be done using mathematical equations or by simulating the system in software.

3. What does the impulse response tell us about a recursive DT system?

The impulse response of a recursive DT system provides information about its stability, causality, and frequency response. It can also be used to determine the system's transfer function, which describes the relationship between the input and output signals.

4. Can the impulse response of a recursive DT system change over time?

Yes, the impulse response of a recursive DT system can change over time if the system's parameters or initial conditions are changed. However, if the system is time-invariant, the impulse response will remain the same.

5. How is the impulse response related to the frequency response of a recursive DT system?

The frequency response of a recursive DT system can be obtained by taking the discrete Fourier transform (DFT) of its impulse response. This allows us to analyze the system's behavior in the frequency domain and determine its frequency response characteristics, such as its magnitude and phase response.

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