Question

3. (30 points) A pressure transducer (a second order system) has natural frequency, fn = 100 kHz, damping ratio 5 = 0.707, and static sensitivity, K=1 mV/psi. (a) Find the (-3 dB) frequency bandwidth few of the transducer. 100 kHz (b) Determine the output of the transducer when used to measure a sinusoidal pressure with amplitude of 5 psi and frequency of 150 kHz. V(t) = 2.03 sin(2 nt 150 10'+-2.10) mV
I see the answers are highlighted, i need an explanation to the answers please.

Answer 1

(a) The system transfer function is given as

                                     

Hie) Vout(s) – Kl_ W Pin(s) 52 +2Ewns + WA

The DC gain of the system is

                                                               

Hs = 0) = K

So -3 dB bandwidth is when

                                  

K = (دز = H ( s | هند + بیع 2 + 2-

Or,

                                                 

ਕ - ਪੰਕ

Or,

                                            

عدد 4 + کند * ارد * 2 دی + ب = 2

But here it is given that

                                                     

{= 0.707 =

So,

                                    

لاند * *4 + 2 * حد * 2 - انت + ب = 2

Or,

                                                         

ابی + ب = 2

Or,

                                                               $\omega^{4} = \omega_{n}^{4}$

Or,

                                                                      

م = اس

So -3 dB bandwidth is given as

                                                           

f = fn = 100 kHz

(b) Input pressure is given by

                                    

Pin(t) = 5 sin(2 * 7 * 150 * 10% *t) psi

Since input has frequency = 150kHz, we need to evaluate the transfer function H(s) at s = j*2*pi*150*10³ so that we get

                                                

H(j2*150*103) = 10-3 (2 * 7 * 105) -(2* 7 * 150 * 103)2 + j2 * 0.707* (2* 7 * 105) * (2* 7 * 150 * 103) + (2* 7 * 105)

Or,

H(j27*150*103) = 10-3 (105) -(150 * 103)2 + j2*0.707 * (105) * (150 * 103) + (105)2

Or,

                            

1010 H(j27 * 150 * 103) = 10-3 -1.25 * 1010+ 2.12 * 1010

Or,

                                   

H(j27 * 150 * 103) = 10-3 1-1.25 + 12.12)

Or,

                                              

H(27 * 150 * 10%) = 4.06 * 10-42 - 120°

When angle is expressed in radian, we get

                                      

H(27 * 150 * 10%) = 4.06 * 10-42 - 2.094 rad

So output voltage is given by

                                              

Vout (t) = 5 * 4.06 * 10-4 sin(2 * 7 * 150 * 10% *t - 2.094) V

Or,

                                              

Vout (t) = 2.03 sin(2 * 7 * 150 * 10% *t - 2.094) mV