Question

Let x(t)=e⁻³ᵗu(t) be a signal, let h(t)=e⁻⁵ᵗu(t) be a linear time invariant system,

Find the mathematival expression of output of the system y(t) by using convolution.

Answer 1

The mathematical expression of the output of the system y(t) by using convolution is y(t) = (1/5) e⁻³ᵗ [e⁵t - 1].

Step-by-step explanation:

To find the mathematical expression of the output of the system y(t) by using convolution, we need to convolve the input signal x(t) with the impulse response of the system h(t). Convolution is a mathematical operation that combines two signals to produce a third signal. In this case, the input signal x(t) is multiplied with a time-reversed and shifted version of the impulse response h(t). The resulting signal y(t) is the output of the system.

Let's use the given signals:

x(t) = e⁻³ᵗu(t)

h(t) = e⁻⁵ᵗu(t)

Now, let's find the convolution of x(t) and h(t). The convolution operation is denoted by asterisk (*).

y(t) = x(t) * h(t)

= ∫[0 to t] x(τ)h(t-τ) dτ

= ∫[0 to t] e⁻³ᵗₐu(tₐ) e⁻⁵(t-tₐ)u(t-tₐ) d(tₐ)

= ∫[0 to t] e⁻³ᵗₐ e⁻⁵(t-tₐ)u(tₐ)u(t-tₐ) d(tₐ)

= ∫[0 to t] e⁻³ᵗₐ e⁻⁵t e⁵tₐu(tₐ)u(t-tₐ) d(tₐ)

= e⁻³ᵗ ∫[0 to t] e⁵tₐu(tₐ)u(t-tₐ) d(tₐ)

= e⁻³ᵗ ∫[0 to t] e⁵tₐ d(tₐ)

= e⁻³ᵗ (1/5) [e⁵tₐ] [0 to t]

= (1/5) e⁻³ᵗ [e⁵t₂ - e⁵t₁]

= (1/5) e⁻³ᵗ [e⁵t - 1]

Therefore, the mathematical expression of the output of the system y(t) by using convolution is y(t) = (1/5) e⁻³ᵗ [e⁵t - 1].