7.6E: Convolution (Exercises)

Q7.6.1

1. Express the inverse transform as an integral.

1. \(1\over s^2(s^2+4)\)
2. \(s\over(s+2)(s^2+9)\)
3. \(s\over(s^2+4)(s^2+9)\)
4. \(s\over(s^2+1)^2\)
5. \(1\over s(s-a)\)
6. \(1\over(s+1)(s^2+2s+2)\)
7. \(1\over (s+1)^2(s^2+4s+5)\)
8. \(1\over(s-1)^3(s+2)^2\)
9. \(s-1\over s^2(s^2-2s+2)\)
10. \(s(s+3)\over(s^2+4)(s^2+6s+10)\)
11. \(1\over(s-3)^5s^6\)
12. \(1\over(s-1)^3(s^2+4)\)
13. \(1\over s^2(s-2)^3\)
14. \(1\over s^7(s-2)^6\)

2. Find the Laplace transform.

1. \(\int_0^t\sin a\tau\cos b(t-\tau)\, d\tau\)
2. \(\int_0^t e^\tau\sin a(t-\tau)\,d\tau\)
3. \(\int_0^t\sinh a\tau\cosh a(t-\tau)\,d\tau\)
4. \(\int_0^t\tau(t-\tau)\sin \omega\tau\cos\omega (t-\tau)\,d\tau\)
5. \(e^t\int_0^t\sin\omega\tau \cos\omega (t-\tau)\,d\tau\)
6. \(e^t\int_0^t\tau^2 (t-\tau)e^\tau\,d\tau\)
7. \(e^{-t}\int_0^t e^{-\tau}\tau\cos\omega (t-\tau)\,d\tau\)
8. \(e^t\int_0^t e^{2\tau}\sinh (t-\tau)\,d\tau\)
9. \(\int_0^t\tau e^{2\tau}\sin 2(t-\tau)\,d\tau\)
10. \(\int_0^t (t-\tau)^3 e^\tau\, d\tau\)
11. \(\int_0^t\tau^6 e^{-(t-\tau)}\sin 3(t-\tau)\,d\tau\)
12. \(\int_0^t\tau^2 (t-\tau)^3\, d\tau\)
13. \(\int_0^t (t-\tau)^7 e^{-\tau} \sin 2\tau\,d\tau\)
14. \(\int_0^t (t-\tau)^4\sin 2\tau\,d\tau\)

3. Find a formula for the solution of the initial value problem.

1. \(y''+3y'+y=f(t),\quad y(0)=0,\quad y'(0)=0\)
2. \(y''+4y=f(t),\quad y(0)=0,\quad y'(0)=0\)
3. \(y''+2y'+y=f(t),\quad y(0)=0,\quad y'(0)=0\)
4. \(y''+k^2y=f(t),\quad y(0)=1,\quad y'(0)=-1\)
5. \(y''+6y'+9y=f(t),\quad y(0)=0,\quad y'(0)=-2\)
6. \(y''-4y=f(t),\quad y(0)=0,\quad y'(0)=3\)
7. \(y''-5y'+6y=f(t),\quad y(0)=1,\quad y'(0)=3\)
8. \(y''+\omega^2y=f(t),\quad y(0)=k_0,\quad y'(0)=k_1\)

4. Solve the integral equation.

1. \(y(t)=t-\int_0^t (t-\tau) y(\tau)\,d\tau\)
2. \(y(t)=\sin t-2 \int_0^t\cos (t-\tau) y (\tau)\,d\tau\)
3. \(y(t)=1+2 \int_0^ty(\tau)\cos(t-\tau)\,d\tau\)
4. \(y(t)=t+\int_0^t y(\tau)e^{-(t-\tau)}\,d\tau\)
5. \(y'(t)=t+\int_0^t y(\tau)\cos (t-\tau)\,d\tau,\, y(0)=4\)
6. \(y(t)=\cos t-\sin t+ \int_0^t y(\tau)\sin (t-\tau)\,d\tau\)

5. Use the convolution theorem to evaluate the integral.

1. \(\int_0^t (t-\tau)^7\tau^8\, d\tau\)
2. \(\int_0^t(t-\tau)^{13}\tau^7\,d\tau\)
3. \(\int_0^t(t-\tau)^6\tau^7\, d\tau\)
4. \(\int_0^te^{-\tau}\sin(t-\tau)\,d\tau\)
5. \(\int_0^t\sin\tau\cos2(t-\tau)\,d\tau\)

6. Show that

\[\int_0^tf(t-\tau)g(\tau)\,d\tau=\int_0^tf(\tau)g(t-\tau)\,d\tau\nonumber \]

by introducing the new variable of integration \(x=t-\tau\) in the first integral.

7. Use the convolution theorem to show that if \(f(t)\leftrightarrow F(s)\) then

\[\int_0^tf(\tau)\,d\tau\leftrightarrow {F(s)\over s}.\nonumber \]

8. Show that if \(p(s)=as^2+bs+c\) has distinct real zeros \(r_1\) and \(r_2\) then the solution of

\[ay''+by'+cy=f(t),\quad y(0)=k_0,\quad y'(0)=k_1\nonumber \]

is

\[\begin{aligned} y(t)&=\; k_0{r_2e^{r_1t}-r_1e^{r_2t}\over r_2-r_1}+k_1{e^{r_2t}-e^{r_1t} \over r_2-r_1} \\ &+{1\over a(r_2-r_1)}\int_0^t(e^{r_2\tau}-e^{r_1\tau})f(t-\tau)\,d\tau.\end{aligned}\nonumber \]

9. Show that if \(p(s)=as^2+bs+c\) has a repeated real zero \(r_1\) then the solution of

\[ay''+by'+cy=f(t),\quad y(0)=k_0,\quad y'(0)=k_1\nonumber \]

is

\[y(t)=\; k_0(1-r_1t)e^{r_1t}+k_1te^{r_1t} +{1\over a}\int_0^t\tau e^{r_1\tau}f(t-\tau)\,d\tau.\nonumber \]

10. Show that if \(p(s)=as^2+bs+c\) has complex conjugate zeros \(\lambda\pm i\omega\) then the solution of

\[ay''+by'+cy=f(t),\quad y(0)=k_0,\quad y'(0)=k_1\nonumber \]

is

\[\begin{aligned} y(t)&=\; e^{\lambda t}\left[k_0(\cos\omega t-{\lambda\over\omega}\sin\omega t)+{k_1\over\omega}\sin\omega t\right] \\ &+{1\over a\omega}\int_0^te^{\lambda t}f(t-\tau)\sin\omega\tau\, d\tau.\end{aligned}\nonumber \]