What are the steps involved in calculating convolution sum

Calculating the convolution sum involves a systematic process that combines two discrete-time signals to produce a third signal. Convolution is a fundamental operation in signal processing and system analysis, and it is essential for understanding how signals interact within a system. Here are the steps involved in calculating the convolution sum:

Convolution Sum Definition

The convolution sum of two discrete-time signals x[n]x[n]x[n] and h[n]h[n]h[n] is given by:

y[n]=(x∗h)[n]=∑k=−∞∞x[k]h[n−k]y[n] = (x * h)[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k]y[n]=(x∗h)[n]=∑k=−∞∞?x[k]h[n−k]

where y[n]y[n]y[n] is the resulting signal, x[n]x[n]x[n] is the input signal, and h[n]h[n]h[n] is the impulse response of the system.

Steps to Calculate the Convolution Sum

1. Understand the Signals:

   • Identify the two discrete-time signals x[n]x[n]x[n] and h[n]h[n]h[n]. Ensure you know their ranges (i.e., the values of nnn for which x[n]x[n]x[n] and h[n]h[n]h[n] are defined).

2. Flip One Signal:

   • Flip one of the signals, typically h[n]h[n]h[n], to obtain h[−k]h[-k]h[−k]. This flipping process involves reversing the time index.

3. Shift the Flipped Signal:

   • Shift the flipped signal h[−k]h[-k]h[−k] by nnn units to get h[n−k]h[n-k]h[n−k]. This step essentially slides the flipped signal along the nnn-axis.

4. Multiply and Sum:

   • For each value of nnn, multiply the corresponding samples of x[k]x[k]x[k] and h[n−k]h[n-k]h[n−k], then sum these products over all values of kkk. This step produces a single value y[n]y[n]y[n] for each nnn.

5. Repeat for All nnn:

   • Repeat the multiplication and summation process for each value of nnn to obtain the complete output signal y[n]y[n]y[n].

Detailed Example

Let's go through an example to illustrate these steps.

Given Signals:

x[n]={1,2,3}x[n] = \{1, 2, 3\}x[n]={1,2,3} h[n]={4,5,6}h[n] = \{4, 5, 6\}h[n]={4,5,6}

Step-by-Step Convolution:

1. Understand the Signals:

   • x[n]x[n]x[n] is defined for n=0,1,2n = 0, 1, 2n=0,1,2.
   • h[n]h[n]h[n] is defined for n=0,1,2n = 0, 1, 2n=0,1,2.

2. Flip h[n]h[n]h[n]:

   • h[−k]h[-k]h[−k] will be {6, 5, 4} (reverse the order of h[n]h[n]h[n]).

3. Shift h[−k]h[-k]h[−k] to Get h[n−k]h[n-k]h[n−k]:

   For each nnn:

   • When n=0n = 0n=0: h[0−k]=h[−k]={6,5,4}h[0-k] = h[-k] = \{6, 5, 4\}h[0−k]=h[−k]={6,5,4}
   • When n=1n = 1n=1: h[1−k]={0,6,5,4}h[1-k] = \{0, 6, 5, 4\}h[1−k]={0,6,5,4} (shift right by 1 position)
   • When n=2n = 2n=2: h[2−k]={0,0,6,5,4}h[2-k] = \{0, 0, 6, 5, 4\}h[2−k]={0,0,6,5,4} (shift right by 2 positions)
   • Continue this process for all values of nnn.

4. Multiply and Sum:

   For each nnn:

   • When n=0n = 0n=0: y[0]=∑k=−∞∞x[k]h[0−k]y[0] = \sum_{k=-\infty}^{\infty} x[k] h[0-k]y[0]=∑k=−∞∞?x[k]h[0−k] y[0]=x[0]⋅h[0]+x[1]⋅h[−1]+x[2]⋅h[−2]y[0] = x[0] \cdot h[0] + x[1] \cdot h[-1] + x[2] \cdot h[-2]y[0]=x[0]⋅h[0]+x[1]⋅h[−1]+x[2]⋅h[−2] y[0]=1⋅6+2⋅5+3⋅4y[0] = 1 \cdot 6 + 2 \cdot 5 + 3 \cdot 4y[0]=1⋅6+2⋅5+3⋅4 y[0]=6+10+12=28y[0] = 6 + 10 + 12 = 28y[0]=6+10+12=28
   • When n=1n = 1n=1: y[1]=∑k=−∞∞x[k]h[1−k]y[1] = \sum_{k=-\infty}^{\infty} x[k] h[1-k]y[1]=∑k=−∞∞?x[k]h[1−k] y[1]=x[0]⋅h[1]+x[1]⋅h[0]+x[2]⋅h[−1]y[1] = x[0] \cdot h[1] + x[1] \cdot h[0] + x[2] \cdot h[-1]y[1]=x[0]⋅h[1]+x[1]⋅h[0]+x[2]⋅h[−1] y[1]=1⋅5+2⋅6+3⋅0y[1] = 1 \cdot 5 + 2 \cdot 6 + 3 \cdot 0y[1]=1⋅5+2⋅6+3⋅0 y[1]=5+12+0=17y[1] = 5 + 12 + 0 = 17y[1]=5+12+0=17
   • When n=2n = 2n=2: y[2]=∑k=−∞∞x[k]h[2−k]y[2] = \sum_{k=-\infty}^{\infty} x[k] h[2-k]y[2]=∑k=−∞∞?x[k]h[2−k] y[2]=x[0]⋅h[2]+x[1]⋅h[1]+x[2]⋅h[0]y[2] = x[0] \cdot h[2] + x[1] \cdot h[1] + x[2] \cdot h[0]y[2]=x[0]⋅h[2]+x[1]⋅h[1]+x[2]⋅h[0] y[2]=1⋅4+2⋅5+3⋅6y[2] = 1 \cdot 4 + 2 \cdot 5 + 3 \cdot 6y[2]=1⋅4+2⋅5+3⋅6 y[2]=4+10+18=32y[2] = 4 + 10 + 18 = 32y[2]=4+10+18=32

5. Complete Output Signal:

   • By continuing this process for all values of nnn, we obtain the full output signal y[n]y[n]y[n].

Summary

The convolution sum y[n]=(x∗h)[n]y[n] = (x * h)[n]y[n]=(x∗h)[n] is computed by:

1. Flipping one of the signals (usually h[n]h[n]h[n]).
2. Shifting the flipped signal.
3. Multiplying the shifted signal with the input signal.
4. Summing the products for each value of nnn.

This process produces the output signal that represents the convolution of the two input signals.