VIPSolutionsgive answer in 2 step with explanation at the end of each step and final answer at the end:Ch) gt est¥e ramet ATR el 1 a Plus Mn Valued fwd coors Of dp now Mater {a crof dp nowmme or LINEREN frithon GOA Lap lat Caleylate Y\ (6) ” le : spliiie | PAPE pal lltreen™ FAIS 1) 1 = 2) Eo Re) c= ar 4) (1— e=2 EE) 5) cos gt Fay ; B 6) sin Gt poy Aber) goo 7) ra cos(F)] TD 8) LL ee) CET S)fems eR EEE] 10) t— 1(i —e— a. bg—bo+c),—(a— BB +ba—c),— (a+ mt bs + By (Pape) (BE) == = 12) e~ tb + (c — bat] Ci 13) eof) sin ft +b cos ot Tm aTR. 3 2. 14) *1— =e) sin Bt + cos pt) Be is = uw? 15) t= le But sin (wav 1-6 t)] Tro From input mulv to output "yl": ¥ I 0103896" 7 fl a iE o- 1932 's + 147864 5 i w Name * Ttf2 = 3 i e 5 i if i ¢ i 3 r
Question:
give answer in 2 step with explanation at the end of each step and final answer at the end:
Ch) gt est¥e ramet
ATR el
1 a
Plus Mn Valued
fwd coors Of dp now Mater
{a crof dp nowmme or
LINEREN frithon GOA
Lap lat
Caleylate Y\ (6)
” le : spliiie |
PAPE pal lltreen™ FAIS
1) 1 =
2) Eo
Re) c= ar
4) (1— e=2 EE)
5) cos gt Fay
; B
6) sin Gt poy
Aber) goo
7) ra cos(F)] TD
8) LL ee) CET
S)fems eR EEE]
10) t— 1(i —e— a.
bg—bo+c),—(a— BB +ba—c),— (a+ mt bs +
By (Pape) (BE) == =
12) e~ tb + (c — bat] Ci
13) eof) sin ft +b cos ot Tm aTR.
3 2.
14) *1— =e) sin Bt + cos pt) Be
is = uw?
15) t= le But sin (wav 1-6 t)] Tro
From input mulv to output "yl": ¥
I 0103896" 7 fl a
iE o- 1932 's + 147864 5 i w
Name * Ttf2 = 3 i e 5 i if i ¢ i 3 r
This is AI generated Response.
Asked by: Test User 100215
Created at: 2025-05-10 18:47:59
Asked by: Test User 100215
Created at: 2025-05-10 18:47:59
Final Answer:
Rendered
Raw Text
Did you like this answer?
Let's solve the Laplace inverse for the transfer function given in the third image:
\[
\frac{0.03896}{s^2 + 0.1932s + 7.786}
\]
We'll break this into **2 steps** as you requested:
---
### **Step 1: Match the form to Laplace pairs**
- The denominator is \( s^2 + 2\alpha s + \omega_n^2 \) where:
- \( 2\alpha = 0.1932 \implies \alpha = 0.0966 \)
- \( \omega_n^2 = 7.786 \implies \omega_n = \sqrt{7.786} \approx 2.791 \)
- The numerator is a constant, so this matches the form of a standard second-order system with zero initial velocity or displacement (impulse response).
From the Laplace table (image 2, pair #15):
\[
\frac{\omega_n}{s^2 + 2\delta\omega_n s + \omega_n^2} \xrightarrow{\mathcal{L}^{-1}} \frac{1}{\sqrt{1-\delta^2}} e^{-\delta\omega_n t} \sin(\omega_n\sqrt{1-\delta^2} t)
\]
Where:
- \( \delta = \frac{\alpha}{\omega_n} = \frac{0.0966}{2.791} \approx 0.0346 \)
**Explanation:**
We have identified the form of the transfer function and matched it with a Laplace pair to prepare for inverse transformation.
---
### **Step 2: Plug in values and write the time-domain solution**
- \( \omega_n \approx 2.791 \)
- \( \delta \approx 0.0346 \)
- The numerator in the Laplace table is \( \omega_n \), but our numerator is \( 0.03896 \), so we need to multiply the time-domain solution by \( \frac{0.03896}{2.791} \).
So,
\[
y(t) = \frac{0.03896}{2.791 \sqrt{1-(0.0346)^2}} e^{-0.0346 \cdot 2.791 t} \sin(2.791\sqrt{1-(0.0346)^2} t)
\]
Calculate the constants:
- \( 2.791\sqrt{1-(0.0346)^2} \approx 2.791 \cdot 0.9994 \approx 2.789 \)
- \( 2.791 \cdot 0.0346 \approx 0.0966 \)
- \( \sqrt{1-0.0346^2} \approx 0.9994 \)
- \( \frac{0.03896}{2.791 \cdot 0.9994} \approx \frac{0.03896}{2.789} \approx 0.01397 \)
So,
\[
\boxed{
y(t) = 0.01397 \; e^{-0.0966 t} \sin(2.789 t)
}
\]
**Explanation:**
We plugged in all the values from the transfer function, simplified, and wrote the final time-domain response using the Laplace transform pair.
---
## **Final Answer:**
\[
\boxed{
y(t) = 0.01397 \; e^{-0.0966 t} \sin(2.789 t)
}
\]
This is the time response for the given transfer function.
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :