Video Transcript - Rose
... A second-order circuit is given in this problem. It has two resistors, one capacitor, and one inductor. Firstly, we want to determine the transfer function, which is the s domain ratio of the output voltage to the input voltage. Let’s convert the circuit into s domain. For a resistor, the impedance ...
... A second-order circuit is given in this problem. It has two resistors, one capacitor, and one inductor. Firstly, we want to determine the transfer function, which is the s domain ratio of the output voltage to the input voltage. Let’s convert the circuit into s domain. For a resistor, the impedance ...
Solved Problems - McMaster University > ECE
... We also have that the critical gain is equal to the Gain Margin Mg . We could thus also use the MATLAB command margin. Solution 5.7. From the definitions of stability margins (see section §5.8 of the book), we know that the gain of the controller can be amplified up to a factor equal to Kc before th ...
... We also have that the critical gain is equal to the Gain Margin Mg . We could thus also use the MATLAB command margin. Solution 5.7. From the definitions of stability margins (see section §5.8 of the book), we know that the gain of the controller can be amplified up to a factor equal to Kc before th ...
( ) R-L-C Circuits and Resonant Circuits
... Note: VR << Vin at this frequency. VR and Vi n are not in phase at this frequency. The little wiggles on VR are real! This behavior is due to the transient solution (homogeneous solution) to the differential eq. describing the circuit. After a few cycles this contribution to VR has died out. ...
... Note: VR << Vin at this frequency. VR and Vi n are not in phase at this frequency. The little wiggles on VR are real! This behavior is due to the transient solution (homogeneous solution) to the differential eq. describing the circuit. After a few cycles this contribution to VR has died out. ...
Lab 7
... Figure 2-1 shows a weighted summer circuit in the inverting configuration. This circuit can be used to sum individual input signals with a variable gain for each signal. The virtual ground at the inverting input terminal of the op-amp keeps the input signals isolated from each other. This isolation ...
... Figure 2-1 shows a weighted summer circuit in the inverting configuration. This circuit can be used to sum individual input signals with a variable gain for each signal. The virtual ground at the inverting input terminal of the op-amp keeps the input signals isolated from each other. This isolation ...
RLC Circuits Note
... A filter can be described by its asymptotic frequency dependence. Although the transfer function may be a complicated complex function of frequency, the asymptotic characteristic is simple. For example, a low-pass filter may have a transfer function that is inversely proportional to frequency in the ...
... A filter can be described by its asymptotic frequency dependence. Although the transfer function may be a complicated complex function of frequency, the asymptotic characteristic is simple. For example, a low-pass filter may have a transfer function that is inversely proportional to frequency in the ...
RLC Circuits Note
... Function Generator and use this sum in your calculations below. Calculate the resonance frequency and measure it by changing the oscillator frequency. Using the measured resonance frequency and resistor value, calculate the Q. Vary the oscillator frequency to find the half-power frequencies and calc ...
... Function Generator and use this sum in your calculations below. Calculate the resonance frequency and measure it by changing the oscillator frequency. Using the measured resonance frequency and resistor value, calculate the Q. Vary the oscillator frequency to find the half-power frequencies and calc ...
Draw the schematic, and label the device sizes.
... There’s a schematic on page 2. Redraw (or cut and paste) that schematic, and identify (circle and label) a. input differential pair (Darlington) b. current mirror active load c. compensation capacitor d. common drain level shifter e. common emitter amplifier (Darlington) f. output stage 2. Assuming ...
... There’s a schematic on page 2. Redraw (or cut and paste) that schematic, and identify (circle and label) a. input differential pair (Darlington) b. current mirror active load c. compensation capacitor d. common drain level shifter e. common emitter amplifier (Darlington) f. output stage 2. Assuming ...
RLC Circuits Note
... A filter can be described by its asymptotic frequency dependence. Although the transfer function may be a complicated complex function of frequency, the asymptotic characteristic is simple. For example, a low-pass filter may have a transfer function that is inversely proportional to frequency in the ...
... A filter can be described by its asymptotic frequency dependence. Although the transfer function may be a complicated complex function of frequency, the asymptotic characteristic is simple. For example, a low-pass filter may have a transfer function that is inversely proportional to frequency in the ...
EE101L Laboratory 5
... RC circuit to sinusoidally varying signals at various frequencies. You will represent these signals in the time and frequency domain and investigate the circuit’s frequency dependence. This frequency dependence can be put to use in analog electronics for designing filters. You will study the basic p ...
... RC circuit to sinusoidally varying signals at various frequencies. You will represent these signals in the time and frequency domain and investigate the circuit’s frequency dependence. This frequency dependence can be put to use in analog electronics for designing filters. You will study the basic p ...
COURSE NUMBER: E E 352 Design of a Low
... filter. After construction we demonstrated that it worked by sending an audio signal from a device through the filter to a low voltage amplifier and speaker. Introduction In this lab we will be analyzing an active low-pass filter circuit. We’ll be examining the theory behind the filter, simulation o ...
... filter. After construction we demonstrated that it worked by sending an audio signal from a device through the filter to a low voltage amplifier and speaker. Introduction In this lab we will be analyzing an active low-pass filter circuit. We’ll be examining the theory behind the filter, simulation o ...
Series_RLC_Circuit
... the reactance of the two components. The reactance, X, depends on the frequency, f: XC = 1/ 2fC and XL = 2fL. t When Xc = XL, the total voltage drop across L and C is zero, since VC o and VL are 180 out of phase and equal in magnitude, cancelling each other out – so VL + VC = 0, and all the voltag ...
... the reactance of the two components. The reactance, X, depends on the frequency, f: XC = 1/ 2fC and XL = 2fL. t When Xc = XL, the total voltage drop across L and C is zero, since VC o and VL are 180 out of phase and equal in magnitude, cancelling each other out – so VL + VC = 0, and all the voltag ...
hw9notready
... a. Calculate and tabulate: i. the overdrive voltage and current in all devices. For this step you may assume that =0. The simplest order may be Mb1 through Mb6, then M1 through M5. ii. Calculate the bias voltages on all nodes, assuming VI,CM=1V. Specifically: tail, G2, G3, G5, G6, S3B, S4AB, and ou ...
... a. Calculate and tabulate: i. the overdrive voltage and current in all devices. For this step you may assume that =0. The simplest order may be Mb1 through Mb6, then M1 through M5. ii. Calculate the bias voltages on all nodes, assuming VI,CM=1V. Specifically: tail, G2, G3, G5, G6, S3B, S4AB, and ou ...
Bode plot
In electrical engineering and control theory, a Bode plot /ˈboʊdi/ is a graph of the frequency response of a system. It is usually a combination of a Bode magnitude plot, expressing the magnitude of the frequency response, and a Bode phase plot, expressing the phase shift. Both quantities are plotted against a horizontal axis proportional to the logarithm of frequency.