Recent content by joel amos

Yes, this is homework. How do I move it? I'd tend to agree, but the question implies that there is. You can see no weaknesses of the taxonomy?

Flynn's Taxonomy was a classification of computer architectures published in 1966. Computing has changed a lot since then. Is this taxonomy still useful? What are some of its shortcomings for evaluating modern computing systems?

Several exercises in my textbook start with assumptions that confuse me. For example: Assume 185 and 122 are signed 8-bit decimal integers stored in sign-magnitude format.Assume 151 and 214 are signed 8-bit decimal integers stored in two's complement format. I am then to go on to find the...

So it should be -15 V?

Well, I don't know how to do the calculation. This graph makes is seem as though they'd be exactly equivalent:

Homework Statement Given the ideal op-amp, determine the resistor R2 and the current I2 where Vin=500mV, Vout = 2.5 V, and R1=5kV. Homework Equations No current passes through op-amps, the op-amp tries to make V+ and V- equal. Ohm's law, voltage divider equation The Attempt at a Solution...

Ah right, so the frequency is simply 4 kHz, which means Vout would be approximately equal to Vin.

(a) I rechecked my work and got 677.3 pF (b) So if without the low pass filter the peak voltage would be 3V. Would the frequency be 2π4000 = 25132.7? What about phase? If I'm assuming the values from part (a), then wouldn't the output voltage match the input voltage as 25.1 kHz < 50kHz and this...

Homework Statement (a) If R1 = 4.7kΩ, what value for C1 will give a cutoff frequency of 50kHz? (b) If the input voltage is described by the equation Vin = 3cos(2π(4000)t), what will the peak output voltage be?[/B] Homework Equations (a) fc = 1/(2πRC) (b) Vout/Vin = 1/√(1 + (2πfRC)2) The...

Okay, that makes sense. Thanks for the response!

Given this ideal op-amp, determine Vout if Vin = 2V and Vs = 15 V. So I know that the ideal op-amp has infinite gain. As the difference between V+ and V- is substantial (2V), wouldn't that mean the output would theoretically by infinitely large but instead max out at Vs? Am I understanding...

Why do we use the square of the graph as opposed to the absolute value of the graph to find the mean values?

I see now that what I was doing was flawed. So 400 Ω is the minimum resistance without allowing more than 30 mA of current. I don't know what the maximum resistance would be. No matter the resistance, the voltage drop over the resistor is 12 V. The LED will never get to its maximum of 75 mW of...

Thanks for being part of the process! This makes a lot of sense now. Thanks again!

Hmm. No current?