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GRE Subject Test · GRE Physics Chủ đề 43 5:26 Giọng đọc tiếng Anh · phụ đề tiếng Anh + 中文 được ghi trực tiếp

chơi không gian · ←/→ 5s · j/l 10s · f toàn màn hình · ,/. tốc độ

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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。

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