Impulse Response of Cascaded Discrete-Time LTI Systems

When two discrete-time linear time-invariant (LTI) systems are connected in cascade, the overall impulse response of the combined system is obtained by convolving the individual impulse responses of the two systems. This means if we have two LTI systems with impulse responses $\text{h}_1\text{[n]}$ and $\text{h}_2\text{[n]}$, the impulse response of the cascaded system, let's call it $\text{h}_{\text{cascade}}\text{[n]}$, will be their convolution:

$$\text{h}_{\text{cascade}}\text{[n]} = \text{h}_1\text{[n]} \ast \text{h}_2\text{[n]}$$

Given Impulse Responses

We are given the impulse responses for the two discrete-time LTI systems:

• System 1: $\text{h}_1\text{[n]} = \delta\text{[n - 1]} + \delta\text{[n + 1]}$
• System 2: $\text{h}_2\text{[n]} = \delta\text{[n]} + \delta\text{[n - 1]}$

Here, $\delta\text{[n]}$ represents the Kronecker delta function, which is $1$ when $\text{n} = 0$ and $0$ otherwise. The property of the Kronecker delta function important for convolution is:

$$\delta\text{[n - a]} \ast \delta\text{[n - b]} = \delta\text{[n - (a + b)]}$$

Calculating the Cascaded Impulse Response

To find the impulse response of the cascaded system, we need to convolve $\text{h}_1\text{[n]}$ and $\text{h}_2\text{[n]}$:

$$\text{h}_{\text{cascade}}\text{[n]} = \left( \delta\text{[n - 1]} + \delta\text{[n + 1]} \right) \ast \left( \delta\text{[n]} + \delta\text{[n - 1]} \right)$$

We can use the distributive property of convolution, similar to multiplication:

$$\text{h}_{\text{cascade}}\text{[n]} = \left( \delta\text{[n - 1]} \ast \delta\text{[n]} \right) + \left( \delta\text{[n - 1]} \ast \delta\text{[n - 1]} \right) + \left( \delta\text{[n + 1]} \ast \delta\text{[n]} \right) + \left( \delta\text{[n + 1]} \ast \delta\text{[n - 1]} \right)$$

Now, let's calculate each term using the property $\delta\text{[n - a]} \ast \delta\text{[n - b]} = \delta\text{[n - (a + b)]}$:

1. First Term: $\delta\text{[n - 1]} \ast \delta\text{[n]}$

   Here, $\text{a} = 1$ and $\text{b} = 0$.

   $$\delta\text{[n - (1 + 0)]} = \delta\text{[n - 1]}$$

2. Second Term: $\delta\text{[n - 1]} \ast \delta\text{[n - 1]}$

   Here, $\text{a} = 1$ and $\text{b} = 1$.

   $$\delta\text{[n - (1 + 1)]} = \delta\text{[n - 2]}$$

3. Third Term: $\delta\text{[n + 1]} \ast \delta\text{[n]}$

   Note that $\delta\text{[n + 1]}$ can be written as $\delta\text{[n - (-1)]}$. So, $\text{a} = -1$ and $\text{b} = 0$.

   $$\delta\text{[n - (-1 + 0)]} = \delta\text{[n - (-1)]} = \delta\text{[n + 1]}$$

4. Fourth Term: $\delta\text{[n + 1]} \ast \delta\text{[n - 1]}$

   Here, $\text{a} = -1$ and $\text{b} = 1$.

   $$\delta\text{[n - (-1 + 1)]} = \delta\text{[n - 0]} = \delta\text{[n]}$$

Now, sum all these individual convolution results to get the overall cascaded impulse response:

$$\text{h}_{\text{cascade}}\text{[n]} = \delta\text{[n - 1]} + \delta\text{[n - 2]} + \delta\text{[n + 1]} + \delta\text{[n]}$$

Rearranging the terms in ascending order of delay (from largest negative shift to largest positive shift):

$$\text{h}_{\text{cascade}}\text{[n]} = \delta\text{[n + 1]} + \delta\text{[n]} + \delta\text{[n - 1]} + \delta\text{[n - 2]}$$

This expression can also be written as:

$$\text{h}_{\text{cascade}}\text{[n]} = \delta\text{[n - 2]} + \delta\text{[n - 1]} + \delta\text{[n]} + \delta\text{[n + 1]}$$

Conclusion

The impulse response of the cascaded system is $\delta\text{[n - 2]} + \delta\text{[n - 1]} + \delta\text{[n]} + \delta\text{[n + 1]}$. This result matches one of the provided options, demonstrating a fundamental concept in discrete-time system analysis.