Recent content by topcat123

Plot it how? Bode, Root locus.

Homework Statement This is the second part my problem. In the first part the Open Loop and Closed loop TF where worked out from design parameters. In later questions a lead or lag commentator is needs to be designed. Design an appropriate controller to ensure the steady state error in the...

Ok \frac{{{\omega}_O}^2}{Q}=\frac{1}{R_1C_1}+\frac{1}{R_1C_2} Giving QQ=\frac{\sqrt{R_1R_2C_1C_2}}{R_1(C_1+C_2)} ??

Thanks I know that {{\omega}_O}^2=\frac{1}{\sqrt{R_1R_2C_1C_2}} so the part in brackets must included Q \left(\frac{1}{R_1C_2}+\frac{1}{R_1C_1}\right)

This is the TF give in a lot of different places for a S & K and as you said it is just a general second order TF. So after reading and researching i have tried the Derive the TE Using Z_1=\frac{1}{j{\omega}C_2}~~~~~~~Z_2=R_1~~~~~~~Z_3=\frac{1}{j{\omega}C_1}~~~~~~~Z_4=R_2 Inserting into the...

I see there are lots of names for the same things The corner frequency or pole frequency wo = undamped natural frequency in rads-1 = wp = pole frequency in rads-1 Q = quality factor = Qp Ok after doing more research this is the TF for a High pass Sallen and Key \frac{V_{out}}{V_{in}}=...

This is the TF given in my learning material is this wrong? I do not know what wp and Qp are abbreviation for. As you can see from above this is the type of format all TF have been derived to by my tutors. Obviously the response is due to the TF. If the circuit has the same basic layout as...

So it is the response of the circuit dictates whether it's a Butterworth or Cherbyshev. Not the circuit design (layout). the TF for the second order S&K \frac{V_{out}}{V_{in}}=\frac{1}{1-\left( \frac{{\omega}_o}{{\omega}} \right)^2-j2\frac{{\omega}}{{\omega}_o}}

Homework Statement Write a report (of approximately 750 words) that discusses and compares the design of each filter in terms of their performance, circuit topology, application and any other relevant characteristics. You may for example attempt to relate the choice of component values to the...

so as far as I figure the TF will be \frac{V_{out}}{V_{in}}=\left(\frac{1}{1+j\frac{{\omega}}{{\omega}_C}}\right)^4 And the phase shift is (-arctan(2{\pi}fRC))*4

Ah I get it the gain is clearly positive +1 and so must the TF And the feedback is negative. Thanks

As the feed back is negative and at unity then 0V potential difference. But I believe there is actual a small amount. Because the gain is large this small difference is amplified so the output become almost equal to non-inverting input voltage at a point of equilibrium.

am ideal op-amp will have infinite input impedance infinite gain and zero output impedance. so the output of the first becomes the input to the second and so on? thanks

I figured it would have -ve feedback as the output feeds back into the inverting input. how do I workout the overall gain with the -80 dB roll off?

Homework Statement 3. FIGURE 3(a) shows a simple low-pass filter followed by a buffer amplifier. (a) Write down the transfer function for the filter. (b) Determine the 3db frequency (fc) if R = 10 kΩ and C = 10 nF. (c) If four such stages are cascaded as shown in FIGURE 3(b), determine the gain...