Systems and signals - integration with dirac's function and step function

Question

I'm taking the introductory course to systems and signals and the mathematics of integration using the step function and Dirac's function

and

For the first integral, I just don't know how to deal with u(t-4), the step function.

The other two questions i don't even know how to start. Thanks, any help would be appreciated

Answer 1

u(t-4) means that the integral is equal to zero till you reach 4, so compute the integral from 4 to infinity. the last question, it's about convolution of h(t) and x(t). for x(t), t<0 always x(t)=0, t=0 x(t) = 1 and t>0 x(t)= -1. draw x(t), you have h(t). convolve them.

Answer 2

I will assume that you want to calculate:

1. The first integral, $$\int_{-\infty}^{\infty}e^{-t^2}tu(t-4)\,dt$$

2. This one: $$y(t)=\int_{-\infty}^{\infty}-(t-\sigma)\mathrm{e}^{-(t-\sigma)}u(t-\sigma)\sigma^2\mathrm{e}^{-\sigma}\sin(\sigma)u(\sigma)\,d\sigma\qquad t\geq0$$

3. And the last one: $$\int_{-\infty}^{\infty}h(t-\tau)x(\tau)\,d\tau$$ with $$x(t)=\delta(t)\mathrm{e}^{-t}-u(t)$$

Let's go step by step.

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FIRST:

$$\int_{-\infty}^{\infty}e^{-t^2}tu(t-4)\,dt$$

Note that the step function u(t-4) means that, before t=4, all the integrand is being multiplied by 0, so: $$\int_{-\infty}^{\infty}e^{-t^2}tu(t-4)\,dt=\int_{4}^{\infty}e^{-t^2}t\,dt$$

Now, notice that $$\frac{d}{dt}\mathrm{e}^{-t^2}=-2t\mathrm{e}^{-t^2}$$ so $$\int_{4}^{\infty}e^{-t^2}t\,dt=-\frac{1}{2}\mathrm{e}^{-t^2}\Bigg|_4^{\infty}=\frac{1}{2}\mathrm{e}^{-16}$$

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SECOND:

Notice that $$ \begin{array}{rcl} y(t)&=&\int_{-\infty}^{\infty}-(t-\sigma)\mathrm{e}^{-(t-\sigma)}u(t-\sigma)\sigma^2\mathrm{e}^{-\sigma}\sin(\sigma)u(\sigma)\,d\sigma\\ &=&\int_{0}^{t}-(t-\sigma)\mathrm{e}^{-(t-\sigma)}\sigma^2\mathrm{e}^{-\sigma}\sin(\sigma)\,d\sigma\\ &=&\mathrm{e}^{-t}\int_{0}^{t}-(t-\sigma)\sigma^2\sin(\sigma)\,d\sigma\\ &=&-t\mathrm{e}^{-t}\int_{0}^{t}\sigma^2\sin(\sigma)\,d\sigma+\mathrm{e}^{-t}\int_{0}^{t}\sigma^3\sin(\sigma)\,d\sigma\\ \end{array}$$

Both integrals can be solved using integration by parts. The first one, let u=sigma^2 and dv=sin(sigma), so $$\begin{array}{rcl} \int_{0}^{t}\sigma^2\sin(\sigma)\,d\sigma&=&-\sigma^2\cos(\sigma)\Bigg|_0^t+\int_0^t\sigma\cos(\sigma)\,d\sigma\\ &=&-\sigma^2\cos(\sigma)\Bigg|_0^t+2\sigma\sin(\sigma)\Bigg|_0^t-2\sigma\Bigg|_0^t\\ &=&-t^2\cos(t)+2t\sin(t)-2t \end{array}$$ You even can use this result to calculate the second integral (I'll leave it to you).

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THIRD: just hints

The important thing here is trying to describe h in terms of step functions. You can do it as: $$h(t)=u(t-1)+u(t-2)-u(t-3)-u(t-4)$$ or, if you want, as rect (boxcar) functions: $$h(t)=\sqcap(t-\frac{3}{2})+2\sqcap(t-\frac{5}{2})+\sqcap(t-\frac{7}{2})$$

If we take the first expression, then $$\begin{array}{rcl} h(t-\tau)x(\tau)&=&\bigg(u(t-\tau-1)+u(t-\tau-2)-u(t-\tau-3)-u(t-\tau-4)\bigg)\\ &&\cdot\bigg(\delta(\tau)\mathrm{e}^{-\tau}-u(\tau)\bigg)\\ &=&\delta(\tau)\mathrm{e}^{-\tau}u(t-\tau-1)-u(\tau)u(t-\tau-1)\\ &&+\delta(\tau)\mathrm{e}^{-\tau}u(t-\tau-2)-u(\tau)u(t-\tau-2)\\ &&-\delta(\tau)\mathrm{e}^{-\tau}u(t-\tau-3)+u(\tau)u(t-\tau-3)\\ &&-\delta(\tau)\mathrm{e}^{-\tau}u(t-\tau-4)+u(\tau)u(t-\tau-4) \end{array}$$

HINT: Note that, if t0 is in the integration interval (a,b), the Dirac delta satisfies $$\int_a^b\delta(t-t_0)f(t)\,dt=f(t_0)$$ Then, note that for the Dirac delta in x(t), t_0=0.