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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 $= t u(t)$

  • $\begingroup$ 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 '19 at 20:15
  • $\begingroup$ also, pretty sure this is a basic math, not a signal processing question. $\endgroup$ – Marcus Müller Mar 21 '19 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 '19 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 '19 at 9:09

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.


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

note :

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


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