Charge de mesure, gain d'amplificateur et réponse du détecteur
GRE Subject Test · GRE Physics Sujet 43 5:26 Narration en anglais · Sous-titres anglais + 中文 incrustés
Chapitres
Transcript
A voltmeter changes the circuit it measures: its reading is the output of a divider, not the open-circuit voltage.
电压表会改变它所测量的电路：它的读数是分压器的输出，而不是开路电压。
This topic calculates meter loading, sets amplifier gain and filter cutoff, and interprets a detector's efficiency, resolution, dead time and calibration.
本主题计算仪表的负载效应，确定放大器增益和滤波器截止频率，并解读探测器的效率、分辨率、死时间和校准。
Model the measured circuit as a Thevenin source V sub th with resistance R sub th and the instrument as input resistance R sub m.
将待测电路建模为戴维南源V_th，其内阻为R_th；仪器输入电阻为R_m。
The reading is V sub th times R sub m, all over R sub m plus R sub th, with the same sign as V sub th and magnitude no greater than the absolute value of V sub th for positive resistances.
读数为V_th乘以R_m除以R_m加R_th，符号与V_th相同，且对于正电阻而言，其幅值不超过|V_th|。
A ten volt source with R sub th equals one hundred kilohms read by a one megohm metre gives ten times one over one point one, which equals nine point zero nine volts, a nine percent low bias.
一个10伏、内阻为100千欧的电源，用1兆欧电表测量时，读数为10乘以1/1.1，即9.09伏，存在9%的低偏误差。
A ten times oscilloscope probe raises the effective input resistance and divides the signal by ten; quote both effects.
10倍示波器探头会提高有效输入电阻，并将信号衰减为原来的十分之一；请同时指出这两种效应。
An ideal non-inverting op-amp stage has gain one plus R2 divided by R1.
理想同相运放级增益为一加R2除以R1。
The inverting stage has negative R2 divided by R1, with the sign carried explicitly.
反相级增益为负R2除以R1，符号显式保留。
A first-order RC low-pass has cutoff f sub c equals one over two pi R C.
一阶RC低通滤波器的截止频率f_c等于1除以2πRC。
Its amplitude ratio is one over the square root of one plus f over f sub c squared.
其幅值比为1除以根号下1加(f/f_c)的平方。
Its phase is negative arctangent of f over f sub c.
其相位为负的arctan(f/f_c)。
At f equals f sub c, the amplitude is one over the square root of two, which is the negative three dB point, not one half.
在f等于f_c时，幅值为1除以根号2，即负3dB点，而非二分之一。
For R equals one kΩ and C equals one hundred nF, f sub c is approximately one point five nine kHz.
当R等于1kΩ且C等于100nF时，f_c约等于1.59kHz。
Detector efficiency epsilon is detected events divided by incident events.
探测器效率 ε 为探测事件数除以入射事件数。
Energy resolution is quoted as FWHM over E, such as 13.2 keV on a 662 keV line, about 2 percent.
能量分辨率以 FWHM 除以 E 表示，例如在 662 keV 谱线上为 13.2 keV，约为 2%。
A non-paralyzable detector with dead time tau records m equals n divided by 1 plus n times tau from true rate n. Inverting gives n equals m divided by 1 minus m times tau, which fails as m times tau approaches 1.
具有死时间 τ 的非瘫痪型探测器记录到的计数率 m 等于真计数率 n 除以 1 加 n 乘 τ，即 m = n / (1 + nτ)；反解得 n = m / (1 − mτ)，当 mτ 趋近于 1 时该式失效。
State which dead-time model you assume.
请说明你所采用的死时间模型。
Paralyzable behaviour differs.
瘫痪型探测器的行为则不同。
A linear calibration against known standards can correct offset and scale error; it does not remove noise or guarantee correction of nonlinearity.
针对已知标准进行线性校准可修正偏移和比例误差，但无法消除噪声或保证修正非线性。
A calibration curve maps indicated to true values; interpolation between points assumes local smoothness.
校准曲线将指示值映射为真实值。 点间插值假设局部平滑。
Report resolution and efficiency separately: sharp peaks with poor efficiency still miss events, and high efficiency with broad resolution still mixes nearby lines.
分辨率与效率应分别报告：尖锐峰形若效率低仍会漏计事件，高灵敏度若分辨率宽仍会使邻近谱线混叠。
A finite voltmeter forms a divider with the source resistance.
有限内阻的电压表与电源内阻构成分压器。
A 12 volt source with 0.2 megohms, read by a 1 megohm meter, gives 12 times 1 over 1.2: 10 volts.
12 伏、内阻 0.2 兆欧的电源，用 1 兆欧的电压表测量，读数为 12 乘 1 除以 1.2：10 伏。
For a non-paralysable detector with observed rate m and dead time tau, the true rate is m over one minus m tau, valid only while m tau is below one.
对于非瘫痪型探测器，观测计数率为 m、死时间为 τ，真实计数率等于 m 除以一减 mτ，只在 mτ 小于一时成立。
Efficiency, dead time and resolution are separate corrections; one cannot replace another.
效率、死时间和分辨率是彼此独立的修正，不能相互替代。
The same source is measured first with a 1 megohm meter, reading 6 volts, and then with a 3 megohm meter, reading 9 volts.
同一电源先用 1 兆欧的电压表测量，读数为 6 伏；再用 3 兆欧的电压表测量，读数为 9 伏。
Find its Thevenin voltage and resistance.
求它的戴维南电压和内阻。
Assume stable linear behaviour.
假设电路保持稳定的线性特性。
Let resistance r be in megohms.
设电阻 r 的单位为兆欧。
The first reading gives V sub th equals 6, times 1 plus r.
第一次读数给出 V_th = 6(1+r)。
The second gives V sub th equals 9, times 3 plus r, all over 3, which simplifies to 9 plus 3r.
第二次读数给出 V_th = 9(3+r)/3 = 9+3r。
Equating yields r equals 1, hence R sub th equals 1 megohm and V sub th equals 12 volts.
联立解得 r = 1，因此 R_th = 1 兆欧，V_th = 12 V。
Both readings are below the unloaded voltage.
两次读数均低于开路电压。
Award two divider equations, resistance and voltage check.
给分点：两个分压方程、电阻值及电压验证。
Treating the one over square root of two cutoff as one half, ignoring that a finite metre reads low through the divider, or applying the non-paralyzable correction beyond its range.
将二分之一根号一的截止频率视为二分之一，忽略有限量程电表通过分压器读数偏低的问题，或在超出其适用范围时应用非瘫痪校正。
dead time: The minimum interval after one recorded event during which the detector cannot record another.
死时间是记录一个事件后，探测器无法记录下一个事件的最短时间间隔。
input impedance: The effective load a measuring instrument presents to the circuit under test.
输入阻抗是测量仪器向被测电路呈现的有效负载。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不阅读定义的情况下回忆其含义，然后改变原始传递函数中的一个参数，并检验极限情况。