Question

The discrete-time system y(n) is the resultant convolution of f(n) and h(n) having lengths 3 and 5 respectively. The maximum possible sample value of f(n) and h(n) are 10 and 20 respectively. Find the maximum possible value of sum of all sample value of y(n).
1. 3000
2. 15
3. 130
4. 200

Answer 1

Correct Answer - Option 1 : 3000

Concept:

If the two discrete signals are having the length ‘n’ and ‘m’ respectively then the resultant output signal has the length as n + m – 1

The convolution of signals in one domain is equivalent to the multiplication of signals in another domain.

Calculation:

Given y[n] = x[n] *h[n]

Operator * denotes the convolution of two signals.

The length of the output signal is y[n] = 3 + 5 – 1 = 7

Y(e^jω) = F(e^jω)H(e^jω)

\(\mathop \sum \limits_{n = 0}^6 y\left[ n \right]{e^{ - j\omega n}} = \mathop \sum \limits_{n = 0}^2 f\left[ n \right]{e^{ - j\omega n}}\mathop \sum \limits_{n = 0}^4 h\left[ n \right]{e^{ - j\omega n}}\)

At ω = 0

\(\mathop \sum \limits_{n = 0}^6 y\left[ n \right] = \mathop \sum \limits_{n = 0}^2 f\left[ n \right]\mathop \sum \limits_{n = 0}^4 h\left[ n \right]\)

 \(max\mathop \sum \limits_{n = 0}^6 y\left[ n \right] = \left( {max\mathop \sum \limits_{n = 0}^2 f\left[ n \right]} \right)\left( {max\mathop \sum \limits_{n = 0}^4 h\left[ n \right]} \right)\)  

\(max\mathop \sum \limits_{n = 0}^6 y\left[ n \right] = \left( {3 \times 10} \right)\left( {5 \times 20} \right)\)

\(max\mathop \sum \limits_{n = 0}^6 y\left[ n \right] = 3000\)