请教高手，两个电路图的输出结果

absolutely Not.
input frequency=?

应该不是吧，你有滤波，怎么可能输入三角波，输出还是三角波?

不是吧
第一个感觉是电压跟随
第二个是电压负反馈 ，然后接低通滤波电路

貌似很难

研究下...

对不起啦， 我是灌水的，下东西，没办法啊哎。

by math
upper circuit
C1 & R3 as high pass pole at 1/(2*PI*1M*4.7u)=3.38e-2Hz~0 -> DC block
R1 & C3 as low pass pole at 1/(2*PI*1K*0.1u)~1.6KHz
OPAmp as non-invert buffer, gain=1
R5&C5 as low pass pole at 1/(2*PI*10K*0.01u)~1.6KHz
So 50Hz saw tooth wave could pass but little round, DC 3V and AC pp 2V is right.

by hand math
lower circuit
C2 & R4 as high pass pole at 1/(2*PI*1M*4.7u)=3.38e-2Hz~0 -> DC block
R2 & C4 as low pass pole at 1/(2*PI*1K*0.1u)~1.6KHz
R7 and R8 and OPAmp as invert amplifier that buffer in/out, gain=1
R6&C6 as low pass pole at 1/(2*PI*10K*0.01u)~1.6KHz
So 50Hz saw tooth wave could pass but little round, DC 3V and AC pp 2V is right.

很簡單，輸入端是一個LPF跟一個HPF，所以就是一個BPF
輸出端也是一個LPF
中間分別是一個BUFFER或一個INVERTING CIRCUIT
按這方向想

其實題目還挺陰險的
你的input波是在0線上震動
可是你的output波在3V線
這是不可動的
除非是同一個Vcm進行
不然應該是不work的

No problems

飘过。，不懂

楼上的几位分析正确