to Figure Q7, determine RE and RB. Given: Av = -8 and re = 5 02. RB CB Figure Q7 20 V 5 ΚΩ не B = 100 ro = ∞0
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- Identify which measurement this oscilloscope time difference reading is useful for. Oscilloscope-XSC1 Channel_B Channel_A 169.064 V 113.590 V -55.474 V Time T1 Reverse 250.000 us 89.282 V 122.375 V 33.093 V T2 382.576 us T2-T1 132.576 us Save Ext. trigger Timebase Scale: 200 us/Div X pos. (Div): 0 Channel A Scale: 100 V/Div Y pos. (Div): 0 Channel B Trigger Scale: 100 V/Div Y pos. (Div): 0 Edge: f 2 AB Ext Level: Y/T Add B/A A/B AC0 DC AC0 DC Single Normal Auto None O none of the above O phase shift O periodIn a frequency modulating system, the frequency deviation constant k = 1 kHz/V. A sinusoidal modulating signal of amplitude is 15 V and a modulated frequency 3 kHz is applied. Calculate the peak frequency deviation Blank 1 kHz and modulation index.Blank 2 Blank 1 Add your answer Blank 2 Add your answer8. Desipn a phanse-lead empensatos s ame k that close-loop poles are at se-tjau5." chane Fhe rerf compansater ot $5-2. Sketeh the unid-s lap sespania of He campesalad aud mucompençaled rysdoms. s(0,55+1) Sheteh the
- Identify which measurement this oscilloscope time difference reading is useful for. Ociloscope-XSC1 Channel_A 169.064 V 168.513 V Channel_B 89.282 V 95.005 V Time Reverse 250.000 us 1.261 ms 1.011 ms T2 T2-T1 -550. 584 mV 5.722 V Save Ext. trigger Channel A Timebase Scale: ko0 us/Div X pos. (Div): 0 Y/T Add B/A A/B Channel B Trigger Scale: 100 V/Div Y pos. (Div): 0 AC0 DC 100 V/Div f 2 AB Ext Scale: Edge: | Y pos.(Div): 0 AC 0 DC Level: V Single Normal Auto NoneQ1/Find (1101100)Gray + (1100111)Ex-3 -(1011011)Gray= ( )16 * None of them O B9 O 75 3F O 4AHere X= 165 this is my roll no plzz solve urgent with detail...
- If A = 2ax + ay - 3az %3D B = ax – az and C = 3ax + 5ay + 7az determine Select your answer. -0.7917a, + 1.0833a, – 0.7917az -1.1667az + 0.7084a, + 0.7084a, 0.7917az – 1.0833a, + 0.7917az 1.1667az – 0.7084ay – 0.7084a,6. Noise power is: * Mark only one oval. P=sqr(k T B) P=k S B P=k T BR V P=k T B 00 0Q.2) Determine all values of resistors and capacitor for the triangular waveform generator whose output alternates between +10Vpeak and -10V peak when the input is a 10-Hz square wave that alternates between +2V and -2V. The input resistance to the generator must be at least 15 kQ. you may assume that there is no de component or offset in the input.
- A clipper is a device that removes either the positive half (top half) or negative half (bottom half), or both positive and negative halves of the input AC signal. In other words, a clipper is a device that limits the positive amplitude or negative amplitude or both positive and negative amplitudes of the input AC signal. In some cases, a clipper removes a small portion of the positive half cycle or negative half cycle or both positive and negative half cycles. In the below circuit diagram, the positive half cycles are removed by using the series positive clipper. QUESTION: In your own opinion. Why do we need to clip a certain amount of voltage in positive or negative or on both sides? What is the benefit of that in our devices or circuits in doing such thing?BDesign a cynchramaus Sequential syncramaus Sequental Circuit for the state diaqram chawn in figure below, using s.R flip.fbp So Sz 0/0 S3Solve for Rc of Figure 42. 12 V Ic Rc RB VCE B=80 IB = 40 μA Reference: Boylestad, R. & Nashehky, L. (2013). Bectronic devices and circuit theory |11th ed.]. Upper Saddle River, New Jersey: Pearson Education Figure 42 1.5 k ohms 1.875 kohms 3.531 kohms 2.75 kohms Vc=6V