0
$\begingroup$

question:

enter image description here

enter image description here

my attempt:

$(a)$ $\displaystyle\int_{-\infty}^{\infty} f(t) dt=2 u (t)+r(t)-r(t-T)+Ku(t-T)$

then we draw it's diagram and found it was unbounded so,we doubt how area can be zero for any value of $K$ so, we need help here finding value of K (which we think will come negative )

$(b)$ $\displaystyle\int_{0_{+}}^{\infty} f(t) dt=r(t)-r(t-T)+Ku(t-T)$

same diagram as part $(a)$ , and same story as above ,except we pull above curve down by two units .

so , we're unable to find value of $K$. any help will be appreciated

thanking you !

edit : $r(t)$ is ramp function $= t u(t)$

$\endgroup$
  • $\begingroup$ hm, you're doing something wrong with your integrals. The solution to these integrals shouldn't depent on $t$, at all. You "integrate the $t$-dependency away", so to speak. $\endgroup$ – Marcus Müller Mar 21 at 20:15
  • $\begingroup$ also, pretty sure this is a basic math, not a signal processing question. $\endgroup$ – Marcus Müller Mar 21 at 20:25
  • $\begingroup$ You're simply doing integration wrong if you still got the $t$-dependency; did you maybe forget to substitute the integral bounds into the antiderivative? I'd recommend to remember how $\delta$ is defined: it's a functional that when integrated over the single point where it's not 0, it has integral value 1; you really don't need to work with the antiderivatives in this exercise. Try simply thinking in areas. $\endgroup$ – Marcus Müller Mar 22 at 7:48
  • $\begingroup$ @Marcus Muller : can you please see answer .given below ...and confirm if it is right or wrong ....i thought in terms of areas as you've said . $\endgroup$ – Faraday Pathak Mar 22 at 9:09
0
$\begingroup$

please someone confirm if following devlopment is right or wrong .becuase answer of exercise question isn't provided in book "F F kuo"

we know

$f(t)\xrightarrow{\mathcal{F}}F(j\omega); $ where, $F(j\omega) =\displaystyle\int_{-\infty}^{\infty} f(t) \ e^{-j\omega t} \ dt\implies F(0)=\text{area under time domain curve }$

So, $(a)\ F(j\omega)=2+T.\ Sa{(\omega T/2)}+Ke^{-j\omega T}\implies F(0)=2+T+K=\displaystyle\int_{-\infty}^{\infty}f(t)dt$

but area is given zero therefore , $K=-(2+T)$ .

and it is resonable also becuase strength of impulse at $t=T$ must be equal to the negative summation of area of regions preceding it.

similarly,

for part $(b)\ $ $K=-T$ because time is starting from $t=0_{+}$ that's why excluding impulse of strength $2$ at $t=0$

note :

$Sa (\omega T/2)=\dfrac{\sin\left(\dfrac{\omega T}{2}\right)}{\omega T/2}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.