(i) 2u(t) with u(t) (2M | May 2015)

(ii) e-2t u(t) with e-5tu(t) (4M | May 2015)

(iii) tu(t) with e-5tu(t) (4M | May 2015)

Subject : Signals & Systems

Topic : Continuous Time Fourier Transform (CTFT) and Discrete Time Fourier Transform (DTFT).

Difficulty: Medium

(i) 2u(t) with u(t)

Let x(t) = 2u(t) and h(t) = u(t)

The output is given by the convolution as

$y(t) =\int_{-∞}^∞ x(τ) h(t- τ) \, dτ $ $y(t) =\int_{-∞}^∞ 2 u(τ) h(t- τ) \, dτ $

u(τ) = 1 for τ ≥ 0

u(t - τ) = 1 for t – τ ≥ 0 i.e. τ ≤ t

The limits of integration will be modified according to above equations i.e. $$y(t) =\int_0^t 2 (1)(1) \, dτ $$

= 2 {[τ]t – [τ]0}

∴y(t) = 2t

(ii) $\ e^{-2t}u(t) \, and \ e^{-5t}u(t) \, $

$Let x(t) = e^{-2t}u(t) \, and \ h(t) = e^{-5t}u(t) \,$

The output is given by the convolution as

$y(t) =\int_{-∞}^∞ x(τ) h(t- τ) \, dτ $

$=\int_{-∞}^∞ e^{-2τ} u(τ) e^{-5(t- τ) } u(t - τ ) \, dτ $

u(τ) = 1 for τ ≥ 0

u(t - τ) = 1 for t – τ ≥ 0 i.e. τ ≤ t

The limits of integration will be modified according to above equations i.e.

$=\int_{-∞}^∞ e^{-2τ} u(τ) e^{-5(t- τ) } u(t - τ ) \, dτ $

$y(t) =\int_0^t e^{-2τ} e^{-5t} e^{5τ} \, dτ $

$y(t) =e^{-5t} \int_0^t e^{3τ} \, dτ $

$=e^{-5t} [ [\frac{e^3τ}3 ]^t - [\frac{e^3τ}3 ]^0]$

$∴y(t) = \frac{e^{-5t}}3 [e^{3t} – 1]$

(iii) t u(t) with $e^{-5t} u(t)$

Let x(t) = t u(t) and $h(t) = e^{-5t} u(t)$

The output is given by the convolution as $y(t) =\int_{-∞}^∞ x(τ) h(t- τ) \, dτ $

$y(t) =\int_{-∞}^∞ τ u(τ) e^{-5(t- τ) } u(t - τ ) \, dτ $

u(τ) = 1 for τ ≥ 0

u(t - τ) = 1 for t – τ ≥ 0 i.e. τ ≤ t

The limits of integration will be modified according to above equations i.e.

$y(t) =\int_{0}^t τ (1) e^{-5(t- τ) } (1) \, dτ $

$y(t) =\int_{0}^t τ e^{-5t} e^{5τ} \, dτ $

$y(t) =\int_{0}^t τe^{5τ} \, dτ $

Integrating using integration by parts

$=e^{-5t} [ [\frac{e^{5τ}}5 ]^t - [\frac{e^{5τ}}5 ]^0] - [ [\frac{e^{5τ}}{25} ]^t - [\frac{e^{5τ}}{25} ]^0]$

$=e^{-5t} [ t[\frac{e^{5τ}-1}3 ] - [\frac{e^{5τ}-1}{25} ]]$

$=e^{-5t} [ [\frac{e^{5τ}t}5 ] - [\frac{t}5 ]] - [ [\frac{e^{5τ}}{25} ]^t - [\frac{1}{25} ]]$

$=e^{-5t} [ [\frac{{(5t-1)}e^{5t}}{25} ] - [\frac{1-5t}{25} ]]$

$= [ [\frac{{(5t-1)}}{25} ] - [\frac{(1-5t)(e^{-5t})}{25} ]]$

$= [ ∴y(t) =[\frac{{(5t-1)-(1-5t) e^{-5t}}}{25} ] $