hmm so the equation must be y(t)=-3+5*sin628t .
What about the resistances??? I tried to solve with the above 3 equations but no luck. Maybe i'm making a mistake during the calculations??
Why is it incorrect? How should I draw the diagram? What do you mean thresholds must be negative? You mean that Vcc in diagram should be -Vcc ?
ω=rad/sec because ω=2*π*f where π=3.14 and f is freq. in Hz.
Φ is in degrees. The waveform diagram starts from 16msec so Φ=0 isnt it?
Hi Eric,
Oh, do you think i gave too much information? He seemed to be having such a hard time solving for the resistor values and even for w so i thought i would help him along. Note i never said what w really was numerically, and i didnt actually solve for any resistor values
I could reduce the information more?
hey. really thanks for all this support
mral i calculated w as you said thats why i used it in the sin. equation 5sin628t. And as i can see you too say there's no Φ (Φ=0).
As soon as i get home, i'll try to solve the whole thing.
eric, i noticed what you said. You mean that these +Vcc values must be negatives, ie -10volts, right?
hey. really thanks for all this support
mral i calculated w as you said thats why i used it in the sin. equation 5sin628t. And as i can see you too say there's no Φ (Φ=0).
As soon as i get home, i'll try to solve the whole thing.
eric, i noticed what you said. You mean that these +Vcc values must be negatives, ie -10volts, right?
Why is this +Vss. How did you input it in the equation?To get the resistor values, we start with the equations similar to yours but we also include the negative supply rail Vss:
D=R1+R2+R3
Vutp=(Vcc-Vss)*(R2+R1)/D+!Vss!
Same here!Vltp=(Vcc-Vss)*(R1)/D+!Vss!
Why's that? How can you say D=100k ?and remember that Vss is negative.
To get the resistance values, we first select a rough impedance for the whole resistance string, say 100k. This means R1+R2+R3=100k.
I rewrote your Vltp equation and i end up here : R1=(Vltp-Vss)*D/(-Vss) , Where Vss=-10 and Vcc=0. Are you sure you didnt make a mistake with the Vss sign of the denominator?Next we solve for R1 from:
Vltp=(Vcc-Vss)*(R1)/D+Vss
and we get:
R1=((Vss-Vltp)*D)/(Vss-Vcc)
I get R2=[Vss*R1+(Vss-Vutp)*D]/Vss , Where Vcc=0, Vss=-10VNow that we know R1 we can solve for R2 from the other equation:
Vutp=(Vcc-Vss)*(R2+R1)/D+Vss
and we get:
R2=((Vss-Vcc)*R1+(Vutp-Vss)*D)/(Vcc-Vss)
Now that we know R2 and R1 we get R3 from knowing D too:
R3=D-R2-R1
and that's about it.
Note that R3 is the resistor that connects to the most positive supply rail Vcc, and R1 is the resistor that connects to the most negative supply rail Vss. You can also do this by taking Vcc=0 (ground) and connect R3 to ground and solving the above with Vcc=0. That means you dont have to use the positive supply rail. Both set points are negative so we can get away with this for this problem.
Also remember that Vss is negative so it enters the equations as a negative number like -5. So for example if Vcc=5v and Vss=-5v then Vcc-Vss=5-(-5)=10, and this is important.
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